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NVAMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

Let ∑n=0∞n3((2n)!)+(2n−1)(n!)(n!)(2n)!=ae+be+c,\sum_{n=0}^{\infty} \frac{n^3 \big( (2n)! \big) + (2n-1)(n!)}{(n!)(2n)!} = a e + \frac{b}{e} + c, where a,b,c∈Za, b, c \in \mathbb{Z} and e=∑n=0∞1n!e = \sum_{n=0}^{\infty} \frac{1}{n!}. Then a2−b+ca^2 - b + c is equal to _____.

Answer

Correct answer:26

Step-by-step solution

Standard Method

Given:

∑n=0∞n3(2n)!+(2n−1)(n!)(n!)(2n)!=ae+be+c\sum_{n=0}^{\infty} \frac{n^3(2n)! + (2n-1)(n!)}{(n!)(2n)!} = ae + \frac{b}{e} + c

and

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

Find: a2−b+ca^2 - b + c.

First simplify the general term:

n3(2n)!+(2n−1)(n!)(n!)(2n)!=n3n!+2n−1(2n)!\frac{n^3(2n)! + (2n-1)(n!)}{(n!)(2n)!} = \frac{n^3}{n!} + \frac{2n-1}{(2n)!}

Hence

S=∑n=0∞n3n!+∑n=0∞2n−1(2n)!S = \sum_{n=0}^{\infty} \frac{n^3}{n!} + \sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}

Now use standard exponential-series identities.

For the first sum,

∑n=0∞n(n−1)(n−2)n!=e,∑n=0∞3n(n−1)n!=3e,∑n=0∞nn!=e\sum_{n=0}^{\infty} \frac{n(n-1)(n-2)}{n!} = e, \qquad \sum_{n=0}^{\infty} \frac{3n(n-1)}{n!} = 3e, \qquad \sum_{n=0}^{\infty} \frac{n}{n!} = e

because

n3=n(n−1)(n−2)+3n(n−1)+nn^3 = n(n-1)(n-2) + 3n(n-1) + n

Therefore

∑n=0∞n3n!=e+3e+e=5e\sum_{n=0}^{\infty} \frac{n^3}{n!} = e + 3e + e = 5e

For the second sum, split it as

∑n=0∞2n−1(2n)!=∑n=0∞2n(2n)!−∑n=0∞1(2n)!\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \sum_{n=0}^{\infty} \frac{2n}{(2n)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}

Now

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

So

∑n=0∞2n−1(2n)!=∑n=1∞1(2n−1)!−∑n=0∞1(2n)!\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}

Using

∑n=0∞1(2n)!=e+e−12,∑n=0∞1(2n+1)!=e−e−12\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{e + e^{-1}}{2}, \qquad \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{e - e^{-1}}{2}

we get

∑n=1∞1(2n−1)!=e−e−12\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{e - e^{-1}}{2}

Hence

∑n=0∞2n−1(2n)!=e−e−12−e+e−12=−1e\sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = \frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2} = -\frac{1}{e}

Therefore

S=5e−1eS = 5e - \frac{1}{e}

Comparing with

S=ae+be+cS = ae + \frac{b}{e} + c

we obtain

a=5,b=−1,c=0a = 5, \qquad b = -1, \qquad c = 0

So

a2−b+c=25−(−1)+0=26a^2 - b + c = 25 - (-1) + 0 = 26

Therefore, the required value is 2626.

The solution is unrelated to this question, so the answer has been derived from the question expression itself.

Common mistakes

  • Treating n3(2n)!+(2n−1)(n!)(n!)(2n)!\frac{n^3(2n)! + (2n-1)(n!)}{(n!)(2n)!} as a single inseparable fraction and not splitting it into n3n!+2n−1(2n)!\frac{n^3}{n!} + \frac{2n-1}{(2n)!}. This hides the standard series forms. Always simplify the term first.

  • Using the wrong identity for ∑n3n!\sum \frac{n^3}{n!}. Writing it directly as ee is incorrect because n3n^3 must be expanded as n(n−1)(n−2)+3n(n−1)+nn(n-1)(n-2) + 3n(n-1) + n before summing.

  • Confusing the even-factorial and odd-factorial sums. ∑n=0∞1(2n)!=e+e−12\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{e + e^{-1}}{2} and ∑n=0∞1(2n+1)!=e−e−12\sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{e - e^{-1}}{2}. Interchanging these gives the wrong sign of the 1e\frac{1}{e} term.

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