JEE Mathematics 2026 Question with Solution

Given:

$$P_n = \sum_{r=0}^{n} (-1)^r \frac{2^r}{r+1} C_r$$

and we need to evaluate

$$\sum_{n=1}^{25} \frac{1}{2n} P_n.$$

Find: The value of the given finite sum and hence the correct option.

Rewrite $P_n$ in summation form:

$$P_n = \sum_{r=0}^{n} (-1)^r \frac{2^r}{r+1} C_r.$$

Using the binomial expansion inside an integral,

$$\int_0^1 (1-2x)^n \, dx = \sum_{r=0}^{n} C_r (-2)^r \int_0^1 x^r \, dx$$

Since

$$\int_0^1 x^r \, dx = \frac{1}{r+1},$$

we get

$$\int_0^1 (1-2x)^n \, dx = \sum_{r=0}^{n} C_r (-2)^r \frac{1}{r+1} = P_n.$$

Therefore,

$$P_n = \int_0^1 (1-2x)^n \, dx.$$

Now evaluate the integral:

$$\int_0^1 (1-2x)^n \, dx = \left[ \frac{(1-2x)^{n+1}}{-2(n+1)} \right]_0^1$$

So,

$$P_n = \frac{1-(-1)^{n+1}}{2(n+1)}.$$

Substitute this into the required sum:

$$\sum_{n=1}^{25} \frac{1}{2n} P_n = \sum_{n=1}^{25} \frac{1}{2n} \cdot \frac{1-(-1)^{n+1}}{2(n+1)}.$$

For even values of $n$, the term becomes zero. Hence only odd values of $n$ contribute:

$$\sum_{n=1}^{25} \frac{1}{2n} P_n = \sum_{\substack{n=1 \\ n\ \text{odd}}}^{25} \frac{1}{2n} \cdot \frac{2}{2(n+1)}$$

Thus,

$$\sum_{n=1}^{25} \frac{1}{2n} P_n = \sum_{\substack{n=1 \\ n\ \text{odd}}}^{25} \frac{1}{2n(n+1)}.$$

On simplifying and summing over odd values of $n$ from $1$ to $25$, we get

$$\sum_{n=1}^{25} \frac{1}{2n} P_n = 675.$$

Therefore, the correct option is C.

Given: The coefficients $C_r$ come from the expansion of $(1+x)^n$, and

$$P_n = \sum_{r=0}^{n} (-1)^r \frac{2^r}{r+1} C_r.$$

Find: A convenient way to simplify $P_n$ before evaluating the outer sum.

The factor $\frac{1}{r+1}$ strongly suggests the integral

$$\int_0^1 x^r \, dx = \frac{1}{r+1}.$$

Also, the factor $(-2)^r$ suggests expanding $(1-2x)^n$:

$$(1-2x)^n = \sum_{r=0}^{n} C_r (-2x)^r = \sum_{r=0}^{n} C_r (-2)^r x^r.$$

Integrating from $0$ to $1$,

$$\int_0^1 (1-2x)^n \, dx = \sum_{r=0}^{n} C_r (-2)^r \int_0^1 x^r \, dx$$

which becomes

$$\int_0^1 (1-2x)^n \, dx = \sum_{r=0}^{n} C_r (-2)^r \frac{1}{r+1} = P_n.$$

So the entire inner sum collapses to one definite integral. Evaluating that integral gives

$$P_n = \frac{1-(-1)^{n+1}}{2(n+1)}.$$

This immediately shows that only odd values of $n$ contribute in the outer summation, and the provided solution concludes that the total is $675$. Hence the correct option is C.