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JEE Mathematics 2025 Question with Solution

The sum 1+1+32!+1+3+53!+1+3+5+74!+…1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + \dots upto ∞\infty terms, is equal to

  • A

    6e6e

  • B

    4e4e

  • C

    3e3e

  • D

    2e2e

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The series is

1+1+32!+1+3+53!+1+3+5+74!+…1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + \dots

Find: Its sum up to ∞\infty terms.

The numerator of the nthn^{\text{th}} term is the sum of the first nn odd numbers:

1+3+5+⋯+(2n−1)=n21 + 3 + 5 + \dots + (2n-1) = n^2

So the general term is

Tn=n2n!T_n = \frac{n^2}{n!}

Hence,

S=∑n=1∞n2n!S = \sum_{n=1}^{\infty} \frac{n^2}{n!}

Now write

n2=n(n−1)+nn^2 = n(n-1) + n

Therefore,

S=∑n=1∞n(n−1)n!+∑n=1∞nn!S = \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!}

For the first sum,

∑n=2∞n(n−1)n!=∑n=2∞1(n−2)!\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} = \sum_{n=2}^{\infty} \frac{1}{(n-2)!}

Let m=n−2m=n-2. Then

∑m=0∞1m!=e\sum_{m=0}^{\infty} \frac{1}{m!} = e

For the second sum,

∑n=1∞nn!=∑n=1∞1(n−1)!\sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!}

Let k=n−1k=n-1. Then

∑k=0∞1k!=e\sum_{k=0}^{\infty} \frac{1}{k!} = e

Therefore,

S=e+e=2eS = e + e = 2e

So, the sum of the given infinite series is 2e2e. The correct option is D.

Using the exponential series idea

Given: S=∑n=1∞n2n!S = \sum_{n=1}^{\infty} \frac{n^2}{n!}

Find: The value of SS.

Use the standard exponential series

e=∑n=0∞1n!e = \sum_{n=0}^{\infty} \frac{1}{n!}

Also,

∑n=1∞nn!=∑n=1∞1(n−1)!=e\sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e

and

∑n=2∞n(n−1)n!=∑n=2∞1(n−2)!=e\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} = \sum_{n=2}^{\infty} \frac{1}{(n-2)!} = e

Since

n2=n(n−1)+nn^2 = n(n-1) + n

we get

∑n=1∞n2n!=∑n=1∞n(n−1)n!+∑n=1∞nn!=e+e=2e\sum_{n=1}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} = e + e = 2e

Therefore, the correct option is D.

Common mistakes

  • Using the numerator as the sum of the first n−1n-1 odd numbers instead of the first nn odd numbers. This shifts the general term incorrectly. Count the odd terms carefully and use 1+3+5+⋯+(2n−1)=n21+3+5+\dots+(2n-1)=n^2.

  • Forgetting to rewrite n2n^2 as n(n−1)+nn(n-1)+n. Without this split, the factorial simplification is not obvious. Break the term in this form so each series reduces to the exponential series.

  • Mismanaging the index shift while converting ∑1(n−2)!\sum \frac{1}{(n-2)!} or ∑1(n−1)!\sum \frac{1}{(n-1)!} into the standard series for ee. After substitution, ensure the new index starts from 00, giving ∑r=0∞1r!=e\sum_{r=0}^{\infty} \frac{1}{r!}=e.

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