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JEE Mathematics 2023 Question with Solution

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The sum ∑n=1∞2n2+3n+4(2n)!\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!} is equal to:

  • A

    11e2+72e\frac{11e}{2} + \frac{7}{2e}

  • B

    13e4+54e−4\frac{13e}{4} + \frac{5}{4e} - 4

  • C

    11e2+72e−4\frac{11e}{2} + \frac{7}{2e} - 4

  • D

    13e4+54e\frac{13e}{4} + \frac{5}{4e}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: ∑n=1∞2n2+3n+4(2n)!\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!}

Find: The value of the infinite sum.

From the extracted working, rewrite the numerator as

2n2+3n+4=2n(2n−1)+8n+82n^2 + 3n + 4 = 2n(2n-1) + 8n + 8

Hence

∑n=1∞2n2+3n+4(2n)!=12∑n=1∞2n(2n−1)(2n)!+2∑n=1∞n(2n−1)!+4∑n=1∞1(2n)!\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{2n(2n-1)}{(2n)!} + 2\sum_{n=1}^{\infty} \frac{n}{(2n-1)!} + 4\sum_{n=1}^{\infty} \frac{1}{(2n)!}

Now,

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

Using the standard even-term exponential expansion,

∑n=1∞1(2n−2)!=e+1e2\sum_{n=1}^{\infty} \frac{1}{(2n-2)!} = \frac{e + \frac{1}{e}}{2}

So the first contribution is

12∑n=1∞2n(2n−1)(2n)!=e+1e4\frac{1}{2}\sum_{n=1}^{\infty} \frac{2n(2n-1)}{(2n)!} = \frac{e + \frac{1}{e}}{4}

Also,

∑n=1∞1(2n−1)!=e−1e2\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{e - \frac{1}{e}}{2}

Therefore, as stated in the extracted solution,

2∑n=1∞n(2n−1)!=e−1e2\sum_{n=1}^{\infty} \frac{n}{(2n-1)!} = e - \frac{1}{e}

And

∑n=1∞1(2n)!=e+1e2−1\sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{e + \frac{1}{e}}{2} - 1

Hence

4∑n=1∞1(2n)!=2(e+1e)−44\sum_{n=1}^{\infty} \frac{1}{(2n)!} = 2\left(e + \frac{1}{e}\right) - 4

Combining all parts,

S=e+1e4+(e−1e)+2(e+1e)−4S = \frac{e + \frac{1}{e}}{4} + \left(e - \frac{1}{e}\right) + 2\left(e + \frac{1}{e}\right) - 4 S=13e4+54e−4S = \frac{13e}{4} + \frac{5}{4e} - 4

Therefore, the sum is 13e4+54e−4\frac{13e}{4} + \frac{5}{4e} - 4 and the correct option is B.

The solution labels option C, but that working belongs to a different question. The answer is therefore resolved from the valid series working provided in the extracted answer text and matched to the options here.

Use even and odd exponential series

Given: ∑n=1∞2n2+3n+4(2n)!\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!}

Find: A faster evaluation using known expansions.

The key idea is to break the numerator into pieces that cancel factorial terms:

2n2+3n+4=2n(2n−1)+8n+82n^2 + 3n + 4 = 2n(2n-1) + 8n + 8

Then each part becomes a standard series involving even or odd factorials. Use

∑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}

After adjusting the starting index n=1n=1, the missing constant term contributes the final −4-4. This is the term most often missed.

Thus the sum simplifies to

13e4+54e−4\frac{13e}{4} + \frac{5}{4e} - 4

So the correct option is B.

Common mistakes

  • Forgetting that ∑n=1∞1(2n)!=e+e−12−1\sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{e + e^{-1}}{2} - 1, not just e+e−12\frac{e + e^{-1}}{2}. This misses the n=0n=0 term and removes the final −4-4. Always adjust for the starting index.

  • Using the wrong even-odd exponential identities. The even factorial series corresponds to e+e−12\frac{e + e^{-1}}{2} and the odd factorial series corresponds to e−e−12\frac{e - e^{-1}}{2}. Interchanging them gives incorrect coefficients of ee and 1e\frac{1}{e}.

  • Decomposing 2n2+3n+42n^2 + 3n + 4 incorrectly. If the numerator is not rewritten in a form that matches factorial cancellation, the series will not reduce cleanly. First create terms like 2n(2n−1)2n(2n-1) so that division by (2n)!(2n)! becomes simpler.

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