JEE Mathematics 2026 Question with Solution

Given: We need the sum of the coefficients of $x^{499}$ and $x^{500}$ in

$$\sum_{k=0}^{1000} x^k(1+x)^{1000-k}$$

Find: The required sum of coefficients and the correct option.

Step 1: Rewrite the expression

$$\sum_{k=0}^{1000} x^k(1+x)^{1000-k}=(1+x)^{1000}\sum_{k=0}^{1000}\left(\frac{x}{1+x}\right)^k$$

Step 2: Evaluate the geometric sum Using

$$\sum_{k=0}^{1000} r^k=\frac{1-r^{1001}}{1-r}, \quad r=\frac{x}{1+x}$$

we get

$$(1+x)^{1000}\cdot\frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}}$$

Since

$$1-\frac{x}{1+x}=\frac{1}{1+x}$$

this becomes

$$(1+x)^{1001}-x^{1001}$$

Step 3: Identify the needed coefficients The term $x^{1001}$ does not affect the coefficients of $x^{499}$ and $x^{500}$. Therefore, we only use $(1+x)^{1001}$.

Coefficient of $x^{499}$ is ${}^{1001}C_{499}$.

Coefficient of $x^{500}$ is ${}^{1001}C_{500}$.

Step 4: Add the coefficients

$${}^{1001}C_{499}+{}^{1001}C_{500} = {}^{1002}C_{500}$$

by Pascal's identity.

Therefore, the correct algebraic result is ${}^{1002}C_{500}$. However, the provided the solution concludes the correct option as D and gives the final answer as ${}^{1002}C_{501}$. Following the solution, the marked correct option is D.

Given:

$$(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000}$$

Find: The sum of the coefficients of $x^{499}$ and $x^{500}$.

Write the series as

$$(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000} =\sum_{k=0}^{1000} x^k(1+x)^{1000-k}$$

Factor out $(1+x)^{1000}$:

$$\sum_{k=0}^{1000} x^k(1+x)^{1000-k} =(1+x)^{1000}\sum_{k=0}^{1000}\left(\frac{x}{1+x}\right)^k$$

Now apply the finite geometric-series formula:

$$\sum_{k=0}^{1000}\left(\frac{x}{1+x}\right)^k =\frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}}$$

Hence

$$(1+x)^{1000}\sum_{k=0}^{1000}\left(\frac{x}{1+x}\right)^k =(1+x)^{1000}\cdot\frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}}$$

Since

$$1-\frac{x}{1+x}=\frac{1}{1+x}$$

we obtain

$$(1+x)^{1000}\cdot\left(1+x\right)\left[1-\left(\frac{x}{1+x}\right)^{1001}\right] =(1+x)^{1001}-x^{1001}$$

Now read the required coefficients from $(1+x)^{1001}$ because $x^{1001}$ does not contribute to powers $x^{499}$ and $x^{500}$.

So the required sum is

$${}^{1001}C_{499}+{}^{1001}C_{500}$$

Using Pascal's identity,

$${}^{1001}C_{499}+{}^{1001}C_{500}={}^{1002}C_{500}$$

the solution, however, lists the correct option as D and writes ${}^{1002}C_{501}$ as the final answer. Thus there is a discrepancy between the displayed working and the stated final option. The extracted official answer from the solution is D.