Simplification
Key Concepts
Solved Examples: 52
Practice Questions: 2
Solved Examples
Question 1: 45 − 18 ÷ 3 × 4 + 12
45 − 18 ÷ 3 × 4 + 12
= 45 − 6 × 4 + 12
= 45 − 24 + 12
= 21 + 12
= 33
= 45 − 6 × 4 + 12
= 45 − 24 + 12
= 21 + 12
= 33
Question 2: 150 ÷ 5 × 2 + 48 ÷ 6 − 9
150 ÷ 5 × 2 + 48 ÷ 6 − 9
= 30 × 2 + 8 − 9
= 60 – 1
= 59
= 30 × 2 + 8 − 9
= 60 – 1
= 59
Question 3: 64 ÷ 8 × 5 − 24 ÷ 6 + 17
64 ÷ 8 × 5 − 24 ÷ 6 + 17
= 8 × 5 − 4 + 17
= 40 + 13
= 53
= 8 × 5 − 4 + 17
= 40 + 13
= 53
Question 4: {72 − (18 × 2)} + (45 ÷ 5)
{72 − (18 × 2)} + (45 ÷ 5)
= {72 − 36} + (9)
= {36} + (9)
= 45
= {72 − 36} + (9)
= {36} + (9)
= 45
Question 5: 8 ÷ 2 × (2 + 2)
8 ÷ 2 × (2 + 2)
Evaluate the bracket first
= 8 ÷ 2 × 4
Perform division and multiplication from left to right.
= 4 x 4
= 16
Evaluate the bracket first
= 8 ÷ 2 × 4
Perform division and multiplication from left to right.
= 4 x 4
= 16
Question 6: 6 ÷ 2 × (1 + 2)
6 ÷ 2 × (1 + 2)
= 6 ÷ 2 × (3)
= 3 x 3
= 9
= 6 ÷ 2 × (3)
= 3 x 3
= 9
Question 7: 96 ÷ 4 × 3 ÷ 6 × 8
96 ÷ 4 × 3 ÷ 6 × 8
Division and multiplication have the same priority, so evaluate them from left to right.
= 24 × 3 ÷ 6 × 8
= 72 ÷ 6 × 8
= 12 x 8
= 96
Division and multiplication have the same priority, so evaluate them from left to right.
= 24 × 3 ÷ 6 × 8
= 72 ÷ 6 × 8
= 12 x 8
= 96
Question 8: If a, b, c, …, x, y, z are 26 consecutive natural numbers, what is the value of (t − a)(t − b)(t − c)…(t − y)(t − z) when t is equal to one of these 26 numbers?
Since t is one of the numbers a, b, c, …, z, one of the factors becomes
(t − t) = 0
A product containing a zero factor is always zero.
Answer = 0
(t − t) = 0
A product containing a zero factor is always zero.
Answer = 0
Question 9: Simplify: (1 − ½)(1 − ⅓)(1 − ¼)…(1 − ¹⁄₂₀)
(1 − ½)(1 − ⅓)(1 − ¼)…(1 − ¹⁄₂₀)
= (½) × (⅔) × (¾) × … × (¹⁹⁄₂₀)
= ¹⁄₂₀
= (½) × (⅔) × (¾) × … × (¹⁹⁄₂₀)
= ¹⁄₂₀
Question 10: If a + 1/b = 1 and b + 1/c = 1, find the value of abc.
From
a + 1/b = 1
a = (b − 1)/b
Similarly,
b = (c − 1)/c
Substitute into the first equation.
a = ((c − 1)/c − 1)/((c − 1)/c)
= −1/(c − 1)
Now,
abc = [−1/(c − 1)] × [(c − 1)/c] × c
= −1
a + 1/b = 1
a = (b − 1)/b
Similarly,
b = (c − 1)/c
Substitute into the first equation.
a = ((c − 1)/c − 1)/((c − 1)/c)
= −1/(c − 1)
Now,
abc = [−1/(c − 1)] × [(c − 1)/c] × c
= −1
Question 11: (1² − 2²) + (2² − 3²) + (3² − 4²) + (4² − 5²)
Expand:
1² − 2² + 2² − 3² + 3² − 4² + 4² − 5²
All middle terms cancel.
= 1² − 5²
= 1 − 25
= −24
1² − 2² + 2² − 3² + 3² − 4² + 4² − 5²
All middle terms cancel.
= 1² − 5²
= 1 − 25
= −24
Question 12: (999 × 1001) − 1000²
Using the identity,
(a − b)(a + b) = a² − b²
(999 × 1001)
= (1000 − 1)(1000 + 1)
= 1000² − 1
Therefore,
(1000² − 1) − 1000²
= −1
(a − b)(a + b) = a² − b²
(999 × 1001)
= (1000 − 1)(1000 + 1)
= 1000² − 1
Therefore,
(1000² − 1) − 1000²
= −1
Question 13: Find the value of 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + 1/(4 × 5)
Use 1/[n(n + 1)] = 1/n − 1/(n + 1)
Therefore,
(1 − ½) + (½ − ⅓) + (⅓ − ¼) + (¼ − ⅕)
Everything cancels.
= 1 − ⅕
= ⁴⁄₅
Therefore,
(1 − ½) + (½ − ⅓) + (⅓ − ¼) + (¼ − ⅕)
Everything cancels.
= 1 − ⅕
= ⁴⁄₅
Question 14: Find the value of: 1 − 2 + 3 − 4 + 5 − 6 + … + 99 − 100
Group the terms in pairs.
(1 − 2) + (3 − 4) + (5 − 6) + … + (99 − 100)
Each pair equals
−1
There are 50 such pairs.
Therefore,
50 × (−1)
= −50
(1 − 2) + (3 − 4) + (5 − 6) + … + (99 − 100)
Each pair equals
−1
There are 50 such pairs.
Therefore,
50 × (−1)
= −50
Question 15: Find the value of (1 + 2 + 3 + … + 50)² − (51 + 52 + … + 100)²
Using the formula Sₙ = n(n + 1)/2
Sum of the first 50 natural numbers:
S₅₀ = 50 × 51 / 2 = 1275
Sum of the first 100 natural numbers:
S₁₀₀ = 100 × 101 / 2 = 5050
Therefore,
51 + 52 + … + 100
= S₁₀₀ − S₅₀
= 5050 − 1275
= 3775
Now,
1275² − 3775²
Using the identity
a² − b² = (a − b)(a + b)
= (1275 − 3775)(1275 + 3775)
= (−2500)(5050)
= −12,625,000
Sum of the first 50 natural numbers:
S₅₀ = 50 × 51 / 2 = 1275
Sum of the first 100 natural numbers:
S₁₀₀ = 100 × 101 / 2 = 5050
Therefore,
51 + 52 + … + 100
= S₁₀₀ − S₅₀
= 5050 − 1275
= 3775
Now,
1275² − 3775²
Using the identity
a² − b² = (a − b)(a + b)
= (1275 − 3775)(1275 + 3775)
= (−2500)(5050)
= −12,625,000
Question 16: Find the missing numerical value represented by the placeholder “?” in the equation: 45 − [38 − {60 ÷ 5 − (? − 3)}] = 19
45 − [38 − {12 − (? − 3)}] = 19
or, 45 − [38 − 12 + (? − 3)] = 19
or, 45 − [26 + (? − 3)] = 19
or, 45 − [23 + ?] = 19
or, 23 + ? = 26
Therefore, ? = 3
or, 45 − [38 − 12 + (? − 3)] = 19
or, 45 − [26 + (? − 3)] = 19
or, 45 − [23 + ?] = 19
or, 23 + ? = 26
Therefore, ? = 3
Question 17: Simplify: 0.5 of 8.4 ÷ 0.07 + 4.2 × 0.5
0.5 × 8.4 ÷ 0.07 + 4.2 × 0.5
= 4.2 ÷ 0.07 + 2.1
= 60 + 2.1
= 62.1
= 4.2 ÷ 0.07 + 2.1
= 60 + 2.1
= 62.1
Question 18: Simplify the sequential division expression: 144 ÷ 12 ÷ 4 ÷ 3
For sequential division, perform the divisions from left to right:
144 ÷ 12 ÷ 4 ÷ 3
= 12 ÷ 4 ÷ 3
= 3 ÷ 3
= 1
144 ÷ 12 ÷ 4 ÷ 3
= 12 ÷ 4 ÷ 3
= 3 ÷ 3
= 1
Question 19: Simplify the continued fraction: 1 + 1 / (1 + 1 / (1 + 1 / 3))
1 + 1 / (1 + 1 / (1 + 1 / 3))
= 1 + 1 / (1 + 1 / (4/3))
= 1 + 1 / (1 + 3/4)
= 1 + 1 / (7/4)
= 1 + 4/7
= 11/7
= 1 + 1 / (1 + 1 / (4/3))
= 1 + 1 / (1 + 3/4)
= 1 + 1 / (7/4)
= 1 + 4/7
= 11/7
Question 20: Simplify the continued fraction: 1 − 1 / (1 − 1 / (1 − 1 / 4))
1 − 1 / (1 − 1 / (1 − 1 / 4))
= 1 − 1 / (1 − 1 / (3/4))
= 1 − 1 / (1 − 4/3)
= 1 − 1 / (−1/3)
= 1 + 3
= 4
= 1 − 1 / (1 − 1 / (3/4))
= 1 − 1 / (1 − 4/3)
= 1 − 1 / (−1/3)
= 1 + 3
= 4
Question 21: Simplify the continued fraction: 1 + 2 / (3 + 4 / (5 + 1 / 2))
1 + 2 / (3 + 4 / (5 + 1 / 2))
= 1 + 2 / (3 + 4 / (11/2))
= 1 + 2 / (3 + 8/11)
= 1 + 2 / (41/11)
= 1 + 22/41
= 63/41
= 1 + 2 / (3 + 4 / (11/2))
= 1 + 2 / (3 + 8/11)
= 1 + 2 / (41/11)
= 1 + 22/41
= 63/41
Question 22: Express the improper fraction 37/16 in the continued fraction form: 2 + 1 / (3 + 1 / ?) and find the value of the missing denominator “?”.
2 + 1 / (3 + 1/?) = 37/16
or, 1 / (3 + 1/?) = 37/16 − 2
or, 1 / (3 + 1/?) = 5/16
or, (3 + 1/?) = 16/5
or, 1/? = 16/5 − 3
or, 1/? = 1/5
or, ? = 5
or, 1 / (3 + 1/?) = 37/16 − 2
or, 1 / (3 + 1/?) = 5/16
or, (3 + 1/?) = 16/5
or, 1/? = 16/5 − 3
or, 1/? = 1/5
or, ? = 5
Question 23: Find the value of: 1/(1 × 4) + 1/(4 × 7) + 1/(7 × 10) + 1/(10 × 13) + 1/(13 × 16)
We can simplify each term using:
1/(n(n + 3)) = 1/3 × [1/n − 1/(n + 3)]
Therefore,
1/(1 × 4) + 1/(4 × 7) + 1/(7 × 10) + 1/(10 × 13) + 1/(13 × 16)
= 1/3 × [(1 − 1/4) + (1/4 − 1/7) + (1/7 − 1/10) + (1/10 − 1/13) + (1/13 − 1/16)]
All the middle terms cancel:
= 1/3 × (1 − 1/16)
= 1/3 × 15/16
= 5/16
1/(n(n + 3)) = 1/3 × [1/n − 1/(n + 3)]
Therefore,
1/(1 × 4) + 1/(4 × 7) + 1/(7 × 10) + 1/(10 × 13) + 1/(13 × 16)
= 1/3 × [(1 − 1/4) + (1/4 − 1/7) + (1/7 − 1/10) + (1/10 − 1/13) + (1/13 − 1/16)]
All the middle terms cancel:
= 1/3 × (1 − 1/16)
= 1/3 × 15/16
= 5/16
Question 24: Find the value of: (1 − 1/2²) × (1 − 1/3²) × (1 − 1/4²) × (1 − 1/5²)
(1 − 1/2²) × (1 − 1/3²) × (1 − 1/4²) × (1 − 1/5²)
= (1 − 1/4) × (1 − 1/9) × (1 − 1/16) × (1 − 1/25)
= 3/4 × 8/9 × 15/16 × 24/25
= 3/5
= (1 − 1/4) × (1 − 1/9) × (1 − 1/16) × (1 − 1/25)
= 3/4 × 8/9 × 15/16 × 24/25
= 3/5
Question 25: Find the value of: 1/(1 × 2 × 3) + 1/(2 × 3 × 4) + 1/(3 × 4 × 5) + 1/(4 × 5 × 6)
1/(1 × 2 × 3) + 1/(2 × 3 × 4) + 1/(3 × 4 × 5) + 1/(4 × 5 × 6)
Use:
1/[n × (n + 1) × (n + 2)] = 1/2 × [1/(n × (n + 1)) − 1/((n + 1) × (n + 2))]
Therefore,
= 1/2 × [1/(1 × 2) − 1/(2 × 3) + 1/(2 × 3) − 1/(3 × 4) + 1/(3 × 4) − 1/(4 × 5) + 1/(4 × 5) − 1/(5 × 6)]
The middle terms cancel:
= 1/2 × [1/2 − 1/30]
= 1/2 × 14/30
= 7/30
Use:
1/[n × (n + 1) × (n + 2)] = 1/2 × [1/(n × (n + 1)) − 1/((n + 1) × (n + 2))]
Therefore,
= 1/2 × [1/(1 × 2) − 1/(2 × 3) + 1/(2 × 3) − 1/(3 × 4) + 1/(3 × 4) − 1/(4 × 5) + 1/(4 × 5) − 1/(5 × 6)]
The middle terms cancel:
= 1/2 × [1/2 − 1/30]
= 1/2 × 14/30
= 7/30
Question 26: Find the value of 100² − 99² + 98² − 97² + 96² − 95² + … + 2² − 1²
100² − 99² + 98² − 97² + 96² − 95² + … + 2² − 1²
Group the terms in pairs:
= (100² − 99²) + (98² − 97²) + … + (2² − 1²)
Using a² − b² = (a − b)(a + b):
= 1 × 199 + 1 × 195 + 1 × 191 + … + 1 × 3
So, = 199 + 195 + 191 + … + 3
This is an arithmetic progression with:
First term = 3
Last term = 199
Common difference = 4
Number of terms = (199 − 3) ÷ 4 + 1 = 50
Therefore,
Sum = 50/2 × (3 + 199)
= 25 × 202
= 5,050
Group the terms in pairs:
= (100² − 99²) + (98² − 97²) + … + (2² − 1²)
Using a² − b² = (a − b)(a + b):
= 1 × 199 + 1 × 195 + 1 × 191 + … + 1 × 3
So, = 199 + 195 + 191 + … + 3
This is an arithmetic progression with:
First term = 3
Last term = 199
Common difference = 4
Number of terms = (199 − 3) ÷ 4 + 1 = 50
Therefore,
Sum = 50/2 × (3 + 199)
= 25 × 202
= 5,050
Question 27: Find the value of the product series: (1 + 1/1) × (1 + 1/2) × (1 + 1/3) × (1 + 1/4) × … × (1 + 1/99)
(1 + 1/1) × (1 + 1/2) × (1 + 1/3) × (1 + 1/4) × … × (1 + 1/99)
= (2/1) × (3/2) × (4/3) × (5/4) × … × (100/99)
Only the very first denominator (1) and the very last numerator (100) remain untouched.
= 100
= (2/1) × (3/2) × (4/3) × (5/4) × … × (100/99)
Only the very first denominator (1) and the very last numerator (100) remain untouched.
= 100
Question 28: Find the value of: 1/(1 × 3) + 1/(3 × 5) + 1/(5 × 7) + … + 1/(19 × 21)
1/(1 × 3) + 1/(3 × 5) + 1/(5 × 7) + … + 1/(19 × 21)
Use: 1/[n(n + 2)] = 1/2 × (1/n − 1/(n + 2))
Therefore,
= 1/2 × [(1/1 − 1/3) + (1/3 − 1/5) + (1/5 − 1/7) + … + (1/19 − 1/21)]
The middle terms cancel:
= 1/2 × (1 − 1/21)
= 1/2 × 20/21
= 10/21
Use: 1/[n(n + 2)] = 1/2 × (1/n − 1/(n + 2))
Therefore,
= 1/2 × [(1/1 − 1/3) + (1/3 − 1/5) + (1/5 − 1/7) + … + (1/19 − 1/21)]
The middle terms cancel:
= 1/2 × (1 − 1/21)
= 1/2 × 20/21
= 10/21
Question 29: Find the value of: (0.87 × 0.87 × 0.87 + 0.13 × 0.13 × 0.13) ÷ (0.87 × 0.87 − 0.87 × 0.13 + 0.13 × 0.13)
Let a = 0.87 and b = 0.13
Then the expression becomes:
(a³ + b³) ÷ (a² − ab + b²)
Using the identity:
a³ + b³ = (a + b)(a² − ab + b²)
Therefore,
(a³ + b³) ÷ (a² − ab + b²)
= (a + b)(a² − ab + b²) ÷ (a² − ab + b²)
= a + b
= 0.87 + 0.13
= 1
Then the expression becomes:
(a³ + b³) ÷ (a² − ab + b²)
Using the identity:
a³ + b³ = (a + b)(a² − ab + b²)
Therefore,
(a³ + b³) ÷ (a² − ab + b²)
= (a + b)(a² − ab + b²) ÷ (a² − ab + b²)
= a + b
= 0.87 + 0.13
= 1
Question 30: Find 3⁄4 of 24% of 800.
3⁄4 × 24% × 800
Convert 24% to a fraction:
= 3⁄4 × 24⁄100 × 800
Simplify:
= 3⁄4 × 24 × 8
= 3 × 6 × 8
= 144
Convert 24% to a fraction:
= 3⁄4 × 24⁄100 × 800
Simplify:
= 3⁄4 × 24 × 8
= 3 × 6 × 8
= 144
Question 31: ?% of 640 − 12.8 = 243.2
?% of 640 − 12.8 = 243.2
Solution
?% of 640 = 243.2 + 12.8
= 256
Therefore,
?% = 256 ÷ 640 × 100
= 40%
Solution
?% of 640 = 243.2 + 12.8
= 256
Therefore,
?% = 256 ÷ 640 × 100
= 40%
Question 32: 27 × 81 ÷ 243 = 3⁽? ⁻ ⁴⁾
27 × 81 ÷ 243 = 3⁽? ⁻ ⁴⁾
or, 3³ × 3⁴ ÷ 3⁵ =3⁽? ⁻ ⁴⁾
or, 3³⁺⁴⁻⁵ = 3⁽? ⁻ ⁴⁾
or, 3² = 3⁽? ⁻ ⁴⁾
or, ? − 4 = 2
or, ? = 6
or, 3³ × 3⁴ ÷ 3⁵ =3⁽? ⁻ ⁴⁾
or, 3³⁺⁴⁻⁵ = 3⁽? ⁻ ⁴⁾
or, 3² = 3⁽? ⁻ ⁴⁾
or, ? − 4 = 2
or, ? = 6
Question 33: 118.3 ÷ 1/0.7 ÷ 13² = 18% of ? + 25% of 7
118.3 ÷ 1/0.7 ÷ 13²
= 118.3 × 0.7 ÷ 169
= 82.81 ÷ 169
= 0.49
Therefore,
18% of ? + 25% of 7 = 0.49
18% of ? + 1.75 = 0.49
18% of ? = −1.26
? = −1.26 × 100 ÷ 18
= −7
= 118.3 × 0.7 ÷ 169
= 82.81 ÷ 169
= 0.49
Therefore,
18% of ? + 25% of 7 = 0.49
18% of ? + 1.75 = 0.49
18% of ? = −1.26
? = −1.26 × 100 ÷ 18
= −7
Question 34: 4.4 + 14.44 + 41.14 = ? − 14.41 − 1.4
4.4 + 14.44 + 41.14 = 59.98
Therefore,
59.98 = ? − 14.41 − 1.4
59.98 = ? − 15.81
? = 59.98 + 15.81
? = 75.79
Therefore,
59.98 = ? − 14.41 − 1.4
59.98 = ? − 15.81
? = 59.98 + 15.81
? = 75.79
Question 35: Find the solution of (1 + 1/3)(1 + 1/3²)(1 + 1/3³)(1 + 1/3⁴) …
We use the identity
1 + x = (1 − x²)/(1 − x)
Taking x = 1/3,
(1 + 1/3)(1 + 1/3²)(1 + 1/3³) …
= (1 − 1/3²)/(1 − 1/3) × (1 − 1/3⁴)/(1 − 1/3²) × (1 − 1/3⁶)/(1 − 1/3⁴) × …
The intermediate terms cancel:
= 1/(1 − 1/3)
= 1/(2/3)
= 3/2
1 + x = (1 − x²)/(1 − x)
Taking x = 1/3,
(1 + 1/3)(1 + 1/3²)(1 + 1/3³) …
= (1 − 1/3²)/(1 − 1/3) × (1 − 1/3⁴)/(1 − 1/3²) × (1 − 1/3⁶)/(1 − 1/3⁴) × …
The intermediate terms cancel:
= 1/(1 − 1/3)
= 1/(2/3)
= 3/2
Question 36: Evaluate the infinite product: (1 − 5⁻¹ + 5⁻²)(1 − 5⁻² + 5⁻⁴)(1 − 5⁻⁴ + 5⁻⁸) …
Let P = (1 − 5⁻¹ + 5⁻²)(1 − 5⁻² + 5⁻⁴)(1 − 5⁻⁴ + 5⁻⁸) …
Multiply both sides by 1 + 5⁻¹ + 5⁻²
Then,
(1 + 5⁻¹ + 5⁻²)(1 − 5⁻¹ + 5⁻²)
= (1 + 5⁻²)² − 5⁻²
= 1 + 5⁻² + 5⁻⁴
Now multiply by the next factor:
(1 + 5⁻² + 5⁻⁴)(1 − 5⁻² + 5⁻⁴)
= (1 + 5⁻⁴)² − 5⁻⁴
= 1 + 5⁻⁴ + 5⁻⁸
Continuing this pattern indefinitely, the powers keep doubling:
(1 + 5⁻¹ + 5⁻²)P = 1 + 0 + 0 = 1
Hence,
P = 1 ÷ (1 + 5⁻¹ + 5⁻²)
= 1 ÷ (1 + 1/5 + 1/25)
= 1 ÷ (31/25)
= 25/31
Multiply both sides by 1 + 5⁻¹ + 5⁻²
Then,
(1 + 5⁻¹ + 5⁻²)(1 − 5⁻¹ + 5⁻²)
= (1 + 5⁻²)² − 5⁻²
= 1 + 5⁻² + 5⁻⁴
Now multiply by the next factor:
(1 + 5⁻² + 5⁻⁴)(1 − 5⁻² + 5⁻⁴)
= (1 + 5⁻⁴)² − 5⁻⁴
= 1 + 5⁻⁴ + 5⁻⁸
Continuing this pattern indefinitely, the powers keep doubling:
(1 + 5⁻¹ + 5⁻²)P = 1 + 0 + 0 = 1
Hence,
P = 1 ÷ (1 + 5⁻¹ + 5⁻²)
= 1 ÷ (1 + 1/5 + 1/25)
= 1 ÷ (31/25)
= 25/31
Question 37: Evaluate (1 − 1/3)(1 + 1/3 + 1/3²)(1 + 1/3³ + 1/3⁶)(1 + 1/3⁹ + 1/3¹⁸) …
We know that (1 − x)(1 + x + x²) = 1 − x³
Consider 1st 2 terms
(1 − 1/3)(1 + 1/3 + 1/3²)
= 1 − 1/3³ ….. new term N1
Then,
N1 x 3rd term
(1 − 1/3³)(1 + 1/3³ + 1/3⁶)
= 1 − 1/3⁹ ….. N2
Continuing indefinitely, the product approaches:
1 − 0 = 1
Consider 1st 2 terms
(1 − 1/3)(1 + 1/3 + 1/3²)
= 1 − 1/3³ ….. new term N1
Then,
N1 x 3rd term
(1 − 1/3³)(1 + 1/3³ + 1/3⁶)
= 1 − 1/3⁹ ….. N2
Continuing indefinitely, the product approaches:
1 − 0 = 1
Question 38: Evaluate (1 − 2⁻¹ + 2⁻²)(1 − 2⁻³ + 2⁻⁶)(1 − 2⁻⁹ + 2⁻¹⁸) …
We know 1 − x + x² = (1 + x³)/(1 + x)
Therefore,
1 − 2⁻¹ + 2⁻² = (1 + 2⁻³)/(1 + 2⁻¹)
Similarly,
1 − 2⁻³ + 2⁻⁶ = (1 + 2⁻⁹)/(1 + 2⁻³)
and so on.
Thus,
P = (1 + 2⁻³)/(1 + 2⁻¹) × (1 + 2⁻⁹)/(1 + 2⁻³) × (1 + 2⁻²⁷)/(1 + 2⁻⁹) …
Everything cancels except the first denominator and the limiting numerator:
P = 1/(1 + 1/2)
= 2/3
Therefore,
1 − 2⁻¹ + 2⁻² = (1 + 2⁻³)/(1 + 2⁻¹)
Similarly,
1 − 2⁻³ + 2⁻⁶ = (1 + 2⁻⁹)/(1 + 2⁻³)
and so on.
Thus,
P = (1 + 2⁻³)/(1 + 2⁻¹) × (1 + 2⁻⁹)/(1 + 2⁻³) × (1 + 2⁻²⁷)/(1 + 2⁻⁹) …
Everything cancels except the first denominator and the limiting numerator:
P = 1/(1 + 1/2)
= 2/3
Question 39: Find the value of 3/(1² × 2²) + 5/(2² × 3²) + 7/(3² × 4²) + … + 19/(9² × 10²)
Notice that each numerator is the difference between the two consecutive squares in its denominator:
2² − 1² = 3
3² − 2² = 5
4² − 3² = 7
and so on.
Therefore, 3/(1² × 2²) = (2² − 1²)/(1² × 2²)
= 1/1² − 1/2²
Similarly, 5/(2² × 3²) = 1/2² − 1/3²
Hence,
3/(1² × 2²) + 5/(2² × 3²) + 7/(3² × 4²) + … + 19/(9² × 10²)
= (1/1² − 1/2²) + (1/2² − 1/3²) + (1/3² − 1/4²) + … + (1/9² − 1/10²)
The intermediate terms cancel:
= 1/1² − 1/10²
= 1 − 1/100
= 99/100
2² − 1² = 3
3² − 2² = 5
4² − 3² = 7
and so on.
Therefore, 3/(1² × 2²) = (2² − 1²)/(1² × 2²)
= 1/1² − 1/2²
Similarly, 5/(2² × 3²) = 1/2² − 1/3²
Hence,
3/(1² × 2²) + 5/(2² × 3²) + 7/(3² × 4²) + … + 19/(9² × 10²)
= (1/1² − 1/2²) + (1/2² − 1/3²) + (1/3² − 1/4²) + … + (1/9² − 1/10²)
The intermediate terms cancel:
= 1/1² − 1/10²
= 1 − 1/100
= 99/100
Question 40: Find the value of 2/(1 × 3) + 4/(3 × 7) + 6/(7 × 13) + 8/(13 × 21)
Notice that the numerator of each fraction is the difference between the two factors in its denominator:
3 − 1 = 2
7 − 3 = 4
13 − 7 = 6
21 − 13 = 8
Therefore,
2/(1 × 3) = (3 − 1)/(1 × 3)
= 1/1 − 1/3
Similarly,
4/(3 × 7) = (7 − 3)/(3 × 7)
= 1/3 − 1/7
6/(7 × 13) = (13 − 7)/(7 × 13)
= 1/7 − 1/13
8/(13 × 21) = (21 − 13)/(13 × 21)
= 1/13 − 1/21
Hence,
2/(1 × 3) + 4/(3 × 7) + 6/(7 × 13) + 8/(13 × 21)
= (1 − 1/3) + (1/3 − 1/7) + (1/7 − 1/13) + (1/13 − 1/21)
The intermediate terms cancel:
= 1 − 1/21
= 20/21
3 − 1 = 2
7 − 3 = 4
13 − 7 = 6
21 − 13 = 8
Therefore,
2/(1 × 3) = (3 − 1)/(1 × 3)
= 1/1 − 1/3
Similarly,
4/(3 × 7) = (7 − 3)/(3 × 7)
= 1/3 − 1/7
6/(7 × 13) = (13 − 7)/(7 × 13)
= 1/7 − 1/13
8/(13 × 21) = (21 − 13)/(13 × 21)
= 1/13 − 1/21
Hence,
2/(1 × 3) + 4/(3 × 7) + 6/(7 × 13) + 8/(13 × 21)
= (1 − 1/3) + (1/3 − 1/7) + (1/7 − 1/13) + (1/13 − 1/21)
The intermediate terms cancel:
= 1 − 1/21
= 20/21
Question 41: Find the value of (1 − 2/(3 × 4)) × (1 − 2/(4 × 5)) × (1 − 2/(5 × 6)) × … × (1 − 2/(11 × 12))
Simplify each factor:
1 − 2/(3 × 4)
= (12 − 2)/12
= 10/12
= (2 × 5)/(3 × 4)
Similarly,
1 − 2/(4 × 5)
= 18/20
= (3 × 6)/(4 × 5)
and
1 − 2/(5 × 6)
= 28/30
= (4 × 7)/(5 × 6)
Therefore,
(1 − 2/(3 × 4)) × (1 − 2/(4 × 5)) × … × (1 − 2/(11 × 12))
= (2 × 5)/(3 × 4) × (3 × 6)/(4 × 5) × (4 × 7)/(5 × 6) × … × (10 × 13)/(11 × 12)
Separate the factors:
= (2/3 × 3/4 × 4/5 × … × 10/11) × (5/4 × 6/5 × 7/6 × … × 13/12)
Now cancel the consecutive factors:
= 2/11 × 13/4
= 26/44
= 13/22
1 − 2/(3 × 4)
= (12 − 2)/12
= 10/12
= (2 × 5)/(3 × 4)
Similarly,
1 − 2/(4 × 5)
= 18/20
= (3 × 6)/(4 × 5)
and
1 − 2/(5 × 6)
= 28/30
= (4 × 7)/(5 × 6)
Therefore,
(1 − 2/(3 × 4)) × (1 − 2/(4 × 5)) × … × (1 − 2/(11 × 12))
= (2 × 5)/(3 × 4) × (3 × 6)/(4 × 5) × (4 × 7)/(5 × 6) × … × (10 × 13)/(11 × 12)
Separate the factors:
= (2/3 × 3/4 × 4/5 × … × 10/11) × (5/4 × 6/5 × 7/6 × … × 13/12)
Now cancel the consecutive factors:
= 2/11 × 13/4
= 26/44
= 13/22
Question 42: Find the value of 1/(1 × 3 × 5) + 1/(3 × 5 × 7) + 1/(5 × 7 × 9) + 1/(7 × 9 × 11)
Notice that the first and last factors in each denominator differ by 4:
5 − 1 = 4
7 − 3 = 4
9 − 5 = 4
11 − 7 = 4
Therefore,
1/(1 × 3 × 5)
= 1/4 × [1/(1 × 3) − 1/(3 × 5)]
Similarly,
1/(3 × 5 × 7)
= 1/4 × [1/(3 × 5) − 1/(5 × 7)]
Hence,
1/(1 × 3 × 5) + 1/(3 × 5 × 7) + 1/(5 × 7 × 9) + 1/(7 × 9 × 11)
= 1/4 × [(1/(1 × 3) − 1/(3 × 5)) + (1/(3 × 5) − 1/(5 × 7)) + (1/(5 × 7) − 1/(7 × 9)) + (1/(7 × 9) − 1/(9 × 11))]
The intermediate terms cancel:
= 1/4 × [1/(1 × 3) − 1/(9 × 11)]
= 1/4 × [1/3 − 1/99]
= 1/4 × [33/99 − 1/99]
= 1/4 × 32/99
= 8/99
5 − 1 = 4
7 − 3 = 4
9 − 5 = 4
11 − 7 = 4
Therefore,
1/(1 × 3 × 5)
= 1/4 × [1/(1 × 3) − 1/(3 × 5)]
Similarly,
1/(3 × 5 × 7)
= 1/4 × [1/(3 × 5) − 1/(5 × 7)]
Hence,
1/(1 × 3 × 5) + 1/(3 × 5 × 7) + 1/(5 × 7 × 9) + 1/(7 × 9 × 11)
= 1/4 × [(1/(1 × 3) − 1/(3 × 5)) + (1/(3 × 5) − 1/(5 × 7)) + (1/(5 × 7) − 1/(7 × 9)) + (1/(7 × 9) − 1/(9 × 11))]
The intermediate terms cancel:
= 1/4 × [1/(1 × 3) − 1/(9 × 11)]
= 1/4 × [1/3 − 1/99]
= 1/4 × [33/99 − 1/99]
= 1/4 × 32/99
= 8/99
Question 43: Find the value of the infinite nested square root √(12 + √(12 + √(12 + …)))
Let the value of the expression be x.
Since the expression repeats indefinitely,
x = √(12 + x)
Squaring both sides:
x² = 12 + x
Therefore,
x² − x − 12 = 0
Factorising:
(x − 4)(x + 3) = 0
Hence, x = 4 or x = −3
Since x represents a square root, it cannot be negative.
Therefore, x = 4
Since the expression repeats indefinitely,
x = √(12 + x)
Squaring both sides:
x² = 12 + x
Therefore,
x² − x − 12 = 0
Factorising:
(x − 4)(x + 3) = 0
Hence, x = 4 or x = −3
Since x represents a square root, it cannot be negative.
Therefore, x = 4
Question 44: Find the value of the product (1 − 1/3) × (1 − 1/6) × (1 − 1/10) × (1 − 1/15) × … × (1 − 1/55)
The denominators 3, 6, 10, 15, …, 55 are triangular numbers:
3 = 1 × 3/2
6 = 2 × 4/2
10 = 3 × 5/2
and so on.
Therefore,
1 − 1/3 = 2/3 = (1 × 4)/(2 × 3)
1 − 1/6 = 5/6 = (2 × 5)/(3 × 4)
1 − 1/10 = 9/10 = (3 × 6)/(4 × 5)
Continuing this pattern,
1 − 1/55 = 54/55 = (9 × 12)/(10 × 11)
Hence,
P = (1 × 4)/(2 × 3) × (2 × 5)/(3 × 4) × (3 × 6)/(4 × 5) × … × (9 × 12)/(10 × 11)
Separate the factors:
P = (1/2 × 2/3 × 3/4 × … × 9/10) × (4/3 × 5/4 × 6/5 × … × 12/11)
Now cancel the consecutive factors:
P = 1/10 × 12/3
= 1/10 × 4
= 2/5
3 = 1 × 3/2
6 = 2 × 4/2
10 = 3 × 5/2
and so on.
Therefore,
1 − 1/3 = 2/3 = (1 × 4)/(2 × 3)
1 − 1/6 = 5/6 = (2 × 5)/(3 × 4)
1 − 1/10 = 9/10 = (3 × 6)/(4 × 5)
Continuing this pattern,
1 − 1/55 = 54/55 = (9 × 12)/(10 × 11)
Hence,
P = (1 × 4)/(2 × 3) × (2 × 5)/(3 × 4) × (3 × 6)/(4 × 5) × … × (9 × 12)/(10 × 11)
Separate the factors:
P = (1/2 × 2/3 × 3/4 × … × 9/10) × (4/3 × 5/4 × 6/5 × … × 12/11)
Now cancel the consecutive factors:
P = 1/10 × 12/3
= 1/10 × 4
= 2/5
Question 45: Find the value of 1/(2 × 4) + 1/(4 × 6) + 1/(6 × 8) + … + 1/(98 × 100)
The difference between the two factors in each denominator is always 2.
We know 1/[n(n + 2)] = 1/2 × (1/n − 1/(n + 2))
we get:
1/(2 × 4) = 1/2 × (1/2 − 1/4)
1/(4 × 6) = 1/2 × (1/4 − 1/6)
Therefore,
1/(2 × 4) + 1/(4 × 6) + 1/(6 × 8) + … + 1/(98 × 100)
= 1/2 × [(1/2 − 1/4) + (1/4 − 1/6) + (1/6 − 1/8) + … + (1/98 − 1/100)]
The intermediate terms cancel:
= 1/2 × (1/2 − 1/100)
= 1/2 × (50/100 − 1/100)
= 1/2 × 49/100
= 49/200
We know 1/[n(n + 2)] = 1/2 × (1/n − 1/(n + 2))
we get:
1/(2 × 4) = 1/2 × (1/2 − 1/4)
1/(4 × 6) = 1/2 × (1/4 − 1/6)
Therefore,
1/(2 × 4) + 1/(4 × 6) + 1/(6 × 8) + … + 1/(98 × 100)
= 1/2 × [(1/2 − 1/4) + (1/4 − 1/6) + (1/6 − 1/8) + … + (1/98 − 1/100)]
The intermediate terms cancel:
= 1/2 × (1/2 − 1/100)
= 1/2 × (50/100 − 1/100)
= 1/2 × 49/100
= 49/200
Question 46: Find the value of the infinite nested square root √(3 × √(3 × √(3 × …)))
Let the value of the expression be x.
Since the expression repeats indefinitely, x = √(3x)
Squaring both sides x² = 3x
Therefore,
x² − 3x = 0
x(x − 3) = 0
Thus,
x = 0 or x = 3
Since the infinite nested expression is positive, x = 3
Since the expression repeats indefinitely, x = √(3x)
Squaring both sides x² = 3x
Therefore,
x² − 3x = 0
x(x − 3) = 0
Thus,
x = 0 or x = 3
Since the infinite nested expression is positive, x = 3
Question 47: Find the value of the infinite nested cube root ∛(16 × ∛(16 × ∛(16 × …)))
Let the value of the entire expression be x.
Then the expression can be written as:
x = ∛(16 × ∛(16 × ∛(16 × …)))
Since the expression continues indefinitely, the part after the first 16 × is again the same expression, x.
Therefore, x = ∛(16 × x)
Cubing both sides x³ = 16x
x³ − 16x = 0
x(x² − 16) = 0
x(x − 4)(x + 4) = 0
Thus, x = 0, 4 or −4
Since the given expression is a positive cube root,
x = 4
Then the expression can be written as:
x = ∛(16 × ∛(16 × ∛(16 × …)))
Since the expression continues indefinitely, the part after the first 16 × is again the same expression, x.
Therefore, x = ∛(16 × x)
Cubing both sides x³ = 16x
x³ − 16x = 0
x(x² − 16) = 0
x(x − 4)(x + 4) = 0
Thus, x = 0, 4 or −4
Since the given expression is a positive cube root,
x = 4
Question 48: Find the value of the infinite nested square root √(24 ÷ √(24 ÷ √(24 ÷ …)))
Let the value of the entire expression be x.
Then, x = √(24 ÷ √(24 ÷ √(24 ÷ …)))
Since the expression continues indefinitely, the portion after the first 24 ÷ is again the same expression, x.
Therefore, x = √(24/x)
Squaring both sides x² = 24/x
Multiplying by x: x³ = 24
Therefore, x = ³√24
Then, x = √(24 ÷ √(24 ÷ √(24 ÷ …)))
Since the expression continues indefinitely, the portion after the first 24 ÷ is again the same expression, x.
Therefore, x = √(24/x)
Squaring both sides x² = 24/x
Multiplying by x: x³ = 24
Therefore, x = ³√24
Question 49: Find the value of the infinite continued fraction 2 + 1/(2 + 1/(2 + 1/(2 + …)))
Let the value of the entire expression be x.
Then, x = 2 + 1/(2 + 1/(2 + 1/(2 + …)))
Since the expression continues indefinitely, the portion after the first 2 + is again the same expression x.
Therefore, x = 2 + 1/x
or, x² = 2x + 1
or, x² − 2x − 1 = 0
or, (x − 1)² = 2
Therefore, x = 1 ± √2
Since the continued fraction is positive and greater than 2,
x = 1 + √2
Then, x = 2 + 1/(2 + 1/(2 + 1/(2 + …)))
Since the expression continues indefinitely, the portion after the first 2 + is again the same expression x.
Therefore, x = 2 + 1/x
or, x² = 2x + 1
or, x² − 2x − 1 = 0
or, (x − 1)² = 2
Therefore, x = 1 ± √2
Since the continued fraction is positive and greater than 2,
x = 1 + √2
Question 50: Find the value of the infinite power tower √2^(√2^(√2^…))
Let the value of the infinite power tower be x.
Since the expression repeats indefinitely,
x = (√2)ˣ
Taking the x-th root of both sides:
x¹ᐟˣ = √2
x¹ᐟˣ = 2¹ᐟ²
Therefore,
1/x = 1/2
Hence, x = 2
Since the expression repeats indefinitely,
x = (√2)ˣ
Taking the x-th root of both sides:
x¹ᐟˣ = √2
x¹ᐟˣ = 2¹ᐟ²
Therefore,
1/x = 1/2
Hence, x = 2
Question 51: Find the value of (1 − 1/3²) × (1 − 1/4²) × (1 − 1/5²) × … × (1 − 1/10²)
Use the identity 1 − 1/n² = (n − 1)(n + 1)/n²
Therefore,
1 − 1/3² = (2 × 4)/3²
1 − 1/4² = (3 × 5)/4²
1 − 1/5² = (4 × 6)/5²
and so on.
Hence,
P = (2 × 4)/3² × (3 × 5)/4² × (4 × 6)/5² × … × (9 × 11)/10²
Separate the factors
P = (2/3 × 3/4 × 4/5 × … × 9/10) × (4/3 × 5/4 × 6/5 × … × 11/10)
Now cancel the consecutive factors:
P = 2/10 × 11/3
= 22/30
= 11/15
Therefore,
1 − 1/3² = (2 × 4)/3²
1 − 1/4² = (3 × 5)/4²
1 − 1/5² = (4 × 6)/5²
and so on.
Hence,
P = (2 × 4)/3² × (3 × 5)/4² × (4 × 6)/5² × … × (9 × 11)/10²
Separate the factors
P = (2/3 × 3/4 × 4/5 × … × 9/10) × (4/3 × 5/4 × 6/5 × … × 11/10)
Now cancel the consecutive factors:
P = 2/10 × 11/3
= 22/30
= 11/15
Question 52: Find the value of the infinite nested radical √(5 × √(2 × √(5 × √(2 × …)))
Let x = √(5 × √(2 × √(5 × √(2 × …))))
The expression immediately inside the first radical has the repeating form:
Let y = √(2 × √(5 × √(2 × …)))
Therefore, x = √(5y) and y = √(2x)
Squaring both equations:
x² = 5y
y² = 2x
From the first equation,
y = x²/5
Substitute this into the second equation:
(x²/5)² = 2x
x⁴/25 = 2x
Since x > 0, divide by x:
x³/25 = 2
Therefore,
x³ = 50
x = ³√50
The expression immediately inside the first radical has the repeating form:
Let y = √(2 × √(5 × √(2 × …)))
Therefore, x = √(5y) and y = √(2x)
Squaring both equations:
x² = 5y
y² = 2x
From the first equation,
y = x²/5
Substitute this into the second equation:
(x²/5)² = 2x
x⁴/25 = 2x
Since x > 0, divide by x:
x³/25 = 2
Therefore,
x³ = 50
x = ³√50
Question 53: Find the value of (1 + 1/2) × (1 − 1/3) × (1 + 1/4) × (1 − 1/5) × … × (1 + 1/50) × (1 − 1/51)
Rewrite each factor:
1 + 1/2 = 3/2
1 − 1/3 = 2/3
1 + 1/4 = 5/4
1 − 1/5 = 4/5
and so on.
Therefore,
P = (3/2) × (2/3) × (5/4) × (4/5) × … × (51/50) × (50/51)
Now observe that the factors cancel in pairs:
(3/2) × (2/3) = 1
(5/4) × (4/5) = 1
and similarly for every subsequent pair.
Hence, P = 1 × 1 × 1 × … × 1 = 1
What is Simplification
Simplification is the process of reducing a mathematical expression to its simplest form by performing mathematical operations in the correct order. It ensures that every expression has a single, correct value.
A simplified expression may involve addition, subtraction, multiplication, division, brackets, exponents, roots, fractions or decimals. To simplify an expression accurately, it is important to follow a standard order of operations.
The most commonly used rule is BODMAS, which determines the sequence in which operations should be carried out.
A simplified expression may involve addition, subtraction, multiplication, division, brackets, exponents, roots, fractions or decimals. To simplify an expression accurately, it is important to follow a standard order of operations.
The most commonly used rule is BODMAS, which determines the sequence in which operations should be carried out.
BODMAS stands for:
B – Brackets
O – Orders (Exponents and Roots)
D – Division
M – Multiplication
A – Addition
S – Subtraction
When multiplication and division appear together, perform the operation that comes first from left to right. The same rule applies to addition and subtraction.
Summary of Simplification
| Concept | Key Point |
|---|---|
| Simplification | Reduce a mathematical expression to its simplest form by following the correct order of operations. |
| Brackets (B) | Solve the expressions inside brackets first. |
| Orders (O) | Evaluate exponents and roots after brackets. |
| Division & Multiplication | Perform division and multiplication from left to right. |
| Addition & Subtraction | Perform addition and subtraction from left to right after all higher-priority operations. |
| Order of Operations: Brackets → Orders → Division → Multiplication → Addition → Subtraction (BODMAS) | |
Common Mistakes
- Ignoring the order of operations: Always follow the BODMAS rule. Solving operations in the wrong order leads to incorrect answers.
- Not working from left to right: When multiplication and division, or addition and subtraction, appear together, solve them from left to right.
- Skipping brackets: Always simplify the innermost brackets before moving to the outer expressions.
- Adding or subtracting fractions directly: Convert fractions to a common denominator before adding or subtracting them.
- Performing addition before multiplication: Multiplication and division must be completed before addition and subtraction unless brackets indicate otherwise.
- Misplacing decimal points: Be careful while multiplying or dividing decimal numbers. A misplaced decimal point can completely change the answer.
- Cancelling terms incorrectly: Cancel only common factors, not terms connected by addition or subtraction.
- Failing to recognise telescoping expressions: Look for patterns where consecutive terms cancel each other. This often leads to a much quicker solution.
Practice Questions
Question 1: {120 − (18 + 24)} ÷ {6 + (18 ÷ 3)} + ⅞ × ⁸⁄₇
Question 2: 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + … + 1/(24 × 25)
Related Topics
 |
|
|
|
|
|