Sequence and Series – Exponential Series – JEE Main 28 Jan 2026 Shift 1

Solution:

Let the general term be $T_k$.
$$T_k = (-1)^{k+1} \frac{k(k+1)}{k!}$$

We can rewrite the numerator $k(k+1)$ as $k(k-1) + 2k$.
$$T_k = (-1)^{k+1} \left[ \frac{k(k-1) + 2k}{k!} \right]$$
$$T_k = (-1)^{k+1} \frac{k(k-1)}{k!} + (-1)^{k+1} \frac{2k}{k!}$$
$$T_k = \frac{(-1)^{k+1}}{(k-2)!} + \frac{2(-1)^{k+1}}{(k-1)!}$$

Now, we sum from $k=1$ to $\infty$. Note that for the first term, the summation effectively starts from $k=2$ because $(k-2)!$ is not defined for $k=1$.
$$S = \sum_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k-1)!}$$

Adjust the indices to match the standard expansion of $e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$.
For the first sum, let $n = k-2$. When $k=2, n=0$. The power of $-1$ is $k+1 = n+3$.
$$\sum_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!} = \sum_{n=0}^{\infty} \frac{(-1)^{n+3}}{n!} = (-1)^3 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = -e^{-1}$$

For the second sum, let $m = k-1$. When $k=1, m=0$. The power of $-1$ is $k+1 = m+2$.
$$2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k-1)!} = 2\sum_{m=0}^{\infty} \frac{(-1)^{m+2}}{m!} = 2(-1)^2 \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} = 2e^{-1}$$

Adding both parts:
$$S = -e^{-1} + 2e^{-1} = e^{-1} = \frac{1}{e}$$

Ans. (2)