JEE Mathematics 2025 Question with Solution

Given:

Find: $\alpha$

Let the left-hand side be $S$. First simplify the term inside the summation:

Now change the index by taking $k = r+1$. Then as $r$ goes from $0$ to $10$, $k$ goes from $1$ to $11$:

Split the summation into two parts:

Use the identity for the sum of binomial coefficients:

So the first part becomes $10(2^{11}-1)$.

For the second part, use the binomial expansion of $\left(1+\frac{1}{10}\right)^{11}$:

Hence the second part becomes

Therefore,

Now equate this with the given right-hand side:

Rewrite the left side with denominator $10^{10}$:

Comparing numerators,

Therefore, the value of $\alpha$ is $20$. The correct option is D.

Given: the solution rewrites the expression into a form involving $^{11}C_{r+1}$ and then compares it with

Find: $\alpha$

The main idea is to express the sum in terms of two standard binomial sums:

1. $\sum {}^{11}C_k$
2. $\sum {}^{11}C_k \left(\frac{1}{10}\right)^k$

After simplification, the sum becomes

Now evaluate each piece separately.

First,

so the first contribution is

Second, from

we get

Multiplying by $10$ gives

Hence,

Comparing this with the right-hand side yields

Therefore, $\alpha = 20$.