Infinite Series MCQ Quiz - Objective Question with Answer for Infinite Series - Download Free PDF

Last updated on Jun 28, 2025

Latest Infinite Series MCQ Objective Questions

Infinite Series Question 1:

Given: The sum of the infinite series: 1/14 + 1/24 + 1/34 + 1/44 + ⋯ = π4 / 90

The sum of the infinite series: 1/14 + 1/34 + 1/54 + ⋯

  1. π4128
  2. π4144
  3. π4120
  4. π496

Answer (Detailed Solution Below)

Option 4 : π496

Infinite Series Question 1 Detailed Solution

Calculation:

We have,

1/14 + 1/24 + 1/34 + 1/44 + ⋯ = π4 / 90         (1)

This can be written as:

[1/14 + 1/34 + 1/54 + ⋯ ] + 1/24 + 1/44 + ⋯ = π4 / 90

[1/14 + 1/34 + 1/54 + ⋯ ] + (1/24) × [1/14 + 1/24 + 1/34 + ⋯ ] = π4 / 90

Using (1):

[1/14 + 1/34 + 1/54 + ⋯ ] + (1/24) × (π4 / 90) = π4 / 90

⇒ 1/14 + 1/34 + 1/54 + ⋯ = (π4 / 90) × [1 − 1/24] = (π4 / 90) × (15/16) = π4 / 96

Infinite Series Question 2:

The value of loge 2 is

  1. 1+12!+13!+14!+
  2. 0
  3. 112+1314+
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 112+1314+

Infinite Series Question 2 Detailed Solution

Concept:

Apply the formula 

loge(1+x)=xx22+x33x44+

Calculation:

We have 

loge(1+x)=xx22+x33x44+

Now put x = 1 then 

loge(1+1)=1122+133144+

loge(2)=112+1314+

Hence the option (3) is correct.

Infinite Series Question 3:

The sum of the series 1+232!+333!+434!+ is

  1. 5e - 1
  2. 5e - 3
  3. 4e
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Infinite Series Question 3 Detailed Solution

Concept:

Expansion of exponential function.

ex=n=0xnn!

=11+x1+x22!+x33!+...

From the above formula value of e1 and e-1 are. 

e1=n=01n!

=11+11!+122!+133!+...

e1=n=0(1)nn!

=1111+12!16+...

Calculation:

Given expansion is:

1+232!+333!+434!+...

we can verify the options and conclude this

Option (A) 5e - 1

5[1+11!+12!+13!+...]1

5+5+52+56+...1

9+52+56+...

not equal to the given series.

option (B) 5e - 3

5[1+11!+12!+13!+...]3

7+52+56+...

not equal to given series.

Option (c) 4e

4[1+11!+12!+13!+...]

4+41!+44!+43!+...

not equal to given series

The answer is option D.

None of the above expansions are matching to given series.

Infinite Series Question 4:

What is the sum to infinity of the series,

3 + 6x2 + 9x4 + 12x6 + ……… given |x| < 1 ?

  1. 3/(1+x2)
  2. 3/(1+x2)2
  3. 3/(1x2)2
  4. 3/(1x2)

Answer (Detailed Solution Below)

Option 3 : 3/(1x2)2

Infinite Series Question 4 Detailed Solution

S(n)=3+6x2+9x4+12x6+....(1)

multiply bu -x2 on both sides

x2S(n)=3x26x49x6...(2)

Adding (1) and (2)

(1x2)S(n)=3+3x2+3x4+3x6+...

(1x2)S(n)=3(1+x2+x4+x6+...)

(1+x2+x4+x6+...)=11x2...(3)

From (3) and (4)

(1x2)S(n)=3×11x2

 S(n)=3×1(1x2)2

Top Infinite Series MCQ Objective Questions

The sum of the series 1+232!+333!+434!+ is

  1. 5e - 1
  2. 5e - 3
  3. 4e
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Infinite Series Question 5 Detailed Solution

Download Solution PDF

Concept:

Expansion of exponential function.

ex=n=0xnn!

=11+x1+x22!+x33!+...

From the above formula value of e1 and e-1 are. 

e1=n=01n!

=11+11!+122!+133!+...

e1=n=0(1)nn!

=1111+12!16+...

Calculation:

Given expansion is:

1+232!+333!+434!+...

we can verify the options and conclude this

Option (A) 5e - 1

5[1+11!+12!+13!+...]1

5+5+52+56+...1

9+52+56+...

not equal to the given series.

option (B) 5e - 3

5[1+11!+12!+13!+...]3

7+52+56+...

not equal to given series.

Option (c) 4e

4[1+11!+12!+13!+...]

4+41!+44!+43!+...

not equal to given series

The answer is option D.

None of the above expansions are matching to given series.

What is the sum to infinity of the series,

3 + 6x2 + 9x4 + 12x6 + ……… given |x| < 1 ?

  1. 3/(1+x2)
  2. 3/(1+x2)2
  3. 3/(1x2)2
  4. 3/(1x2)

Answer (Detailed Solution Below)

Option 3 : 3/(1x2)2

Infinite Series Question 6 Detailed Solution

Download Solution PDF

S(n)=3+6x2+9x4+12x6+....(1)

multiply bu -x2 on both sides

x2S(n)=3x26x49x6...(2)

Adding (1) and (2)

(1x2)S(n)=3+3x2+3x4+3x6+...

(1x2)S(n)=3(1+x2+x4+x6+...)

(1+x2+x4+x6+...)=11x2...(3)

From (3) and (4)

(1x2)S(n)=3×11x2

 S(n)=3×1(1x2)2

Infinite Series Question 7:

The sum of the series 1+232!+333!+434!+ is

  1. 5e - 1
  2. 5e - 3
  3. 4e
  4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Infinite Series Question 7 Detailed Solution

Concept:

Expansion of exponential function.

ex=n=0xnn!

=11+x1+x22!+x33!+...

From the above formula value of e1 and e-1 are. 

e1=n=01n!

=11+11!+122!+133!+...

e1=n=0(1)nn!

=1111+12!16+...

Calculation:

Given expansion is:

1+232!+333!+434!+...

we can verify the options and conclude this

Option (A) 5e - 1

5[1+11!+12!+13!+...]1

5+5+52+56+...1

9+52+56+...

not equal to the given series.

option (B) 5e - 3

5[1+11!+12!+13!+...]3

7+52+56+...

not equal to given series.

Option (c) 4e

4[1+11!+12!+13!+...]

4+41!+44!+43!+...

not equal to given series

The answer is option D.

None of the above expansions are matching to given series.

Infinite Series Question 8:

The value of loge 2 is

  1. 1+12!+13!+14!+
  2. 0
  3. 112+1314+
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 112+1314+

Infinite Series Question 8 Detailed Solution

Concept:

Apply the formula 

loge(1+x)=xx22+x33x44+

Calculation:

We have 

loge(1+x)=xx22+x33x44+

Now put x = 1 then 

loge(1+1)=1122+133144+

loge(2)=112+1314+

Hence the option (3) is correct.

Infinite Series Question 9:

What is the sum to infinity of the series,

3 + 6x2 + 9x4 + 12x6 + ……… given |x| < 1 ?

  1. 3/(1+x2)
  2. 3/(1+x2)2
  3. 3/(1x2)2
  4. 3/(1x2)

Answer (Detailed Solution Below)

Option 3 : 3/(1x2)2

Infinite Series Question 9 Detailed Solution

S(n)=3+6x2+9x4+12x6+....(1)

multiply bu -x2 on both sides

x2S(n)=3x26x49x6...(2)

Adding (1) and (2)

(1x2)S(n)=3+3x2+3x4+3x6+...

(1x2)S(n)=3(1+x2+x4+x6+...)

(1+x2+x4+x6+...)=11x2...(3)

From (3) and (4)

(1x2)S(n)=3×11x2

 S(n)=3×1(1x2)2

Infinite Series Question 10:

Given: The sum of the infinite series: 1/14 + 1/24 + 1/34 + 1/44 + ⋯ = π4 / 90

The sum of the infinite series: 1/14 + 1/34 + 1/54 + ⋯

  1. π4128
  2. π4144
  3. π4120
  4. π496

Answer (Detailed Solution Below)

Option 4 : π496

Infinite Series Question 10 Detailed Solution

Calculation:

We have,

1/14 + 1/24 + 1/34 + 1/44 + ⋯ = π4 / 90         (1)

This can be written as:

[1/14 + 1/34 + 1/54 + ⋯ ] + 1/24 + 1/44 + ⋯ = π4 / 90

[1/14 + 1/34 + 1/54 + ⋯ ] + (1/24) × [1/14 + 1/24 + 1/34 + ⋯ ] = π4 / 90

Using (1):

[1/14 + 1/34 + 1/54 + ⋯ ] + (1/24) × (π4 / 90) = π4 / 90

⇒ 1/14 + 1/34 + 1/54 + ⋯ = (π4 / 90) × [1 − 1/24] = (π4 / 90) × (15/16) = π4 / 96

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