The sum 1 + (1+3)/2! + (1+3+5)/3! + (1+3+5+7)/4! + ... upto n terms, is equal to

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q6 – JEE-MAIN 2025

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Home Maths Sequences and SeriesQuestion #409

The sum

1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .

upto

n

terms, is equal to

Clue 1 of 3: Identify the general term

Observe the pattern in the numerator and denominator of each term.

🥷

Ninja Strategy•Magnitude Estimation

Calculate the sum of the first few terms of the series and compare it to the approximate values of the options to eliminate those that are clearly too large.

🧠 Full Solution Path

Step 1: Identify the General Term✦ Active

The given series is $S = 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .$ . The $k$ -th term of the series, $T_{k}$ , has a numerator which is the sum of the first $k$ odd numbers, which is $k^{2}$ . The denominator is $k!$ . Thus, the general term is $T_{k} = \frac{k^{2}}{k!}$ .

S = \sum_{k = 1}^{\infty} \frac{k^{2}}{k!}

Step 2: Manipulate the General Term○ Expand

We can rewrite the general term $T_{k}$ as follows:

\frac{k^{2}}{k!} = \frac{k}{(k - 1)!} = \frac{(k - 1) + 1}{(k - 1)!} = \frac{k - 1}{(k - 1)!} + \frac{1}{(k - 1)!}

For $k \geq 2$ , $\frac{k - 1}{(k - 1)!} = \frac{1}{(k - 2)!}$ . For $k = 1$ , $\frac{1 - 1}{(1 - 1)!} = \frac{0}{0!} = 0$ . So, the term can be written as:

T_{k} = \frac{1}{(k - 2)!} + \frac{1}{(k - 1)!} for k \geq 2

For $k = 1$ , $T_{1} = \frac{1^{2}}{1!} = 1$ . The formula $\frac{1}{(k - 2)!} + \frac{1}{(k - 1)!}$ for $k = 1$ would involve negative factorials, so we handle $T_{1}$ separately.

💡 Teacher's Secret Hint

Remember to handle the $k = 1$ term carefully when simplifying factorial expressions.

Step 3: Sum the Series using Exponential Series○ Expand

The sum $S$ can be written as:

S = T_{1} + \sum_{k = 2}^{\infty} (\frac{1}{(k - 2)!} + \frac{1}{(k - 1)!})

Substitute $T_{1} = 1$ and split the sum:

S = 1 + \sum_{k = 2}^{\infty} \frac{1}{(k - 2)!} + \sum_{k = 2}^{\infty} \frac{1}{(k - 1)!}

Let $j = k - 2$ in the first sum. When $k = 2$ , $j = 0$ . So, $\sum_{k = 2}^{\infty} \frac{1}{(k - 2)!} = \sum_{j = 0}^{\infty} \frac{1}{j!} = e$ (the definition of $e$ ).

Let $m = k - 1$ in the second sum. When $k = 2$ , $m = 1$ . So, $\sum_{k = 2}^{\infty} \frac{1}{(k - 1)!} = \sum_{m = 1}^{\infty} \frac{1}{m!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . . = e - \frac{1}{0!} = e - 1$ .

Substitute these back into the expression for $S$ :

S = 1 + e + (e - 1) = 2 e

💡 Teacher's Secret Hint

Recall the Taylor series expansion for $e^{x}$ and specifically for $e$ when $x = 1$ .

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Maths Question 6 – JEE-MAIN 2025

Clue 1 of 3: Identify the general term

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