JEE Mathematics 2024 Question with Solution

Given:

$$S = x^2(1 + x)^{98} + x^3(1 + x)^{97} + x^4(1 + x)^{96} + \ldots + x^{54}(1 + x)^{46}$$

Find: The possible value of $p+q$ when the coefficient of $x^{70}$ is $^{99}C_p - {}^{46}C_q$.

The terms form a geometric progression. Therefore,

$$S = x^2(1 + x)^{98} \left[ \frac{\left( \frac{x}{1 + x} \right)^{53} - 1}{\frac{x}{1 + x} - 1} \right]$$

Now,

$$S = x^2(1 + x)^{46} \left[ (1 + x)^{53} - x^{53} \right]$$

So,

$$S = x^2(1 + x)^{99} - x^{55}(1 + x)^{46}$$

The coefficient of $x^{70}$ is therefore

$${}^{99}C_{68} - {}^{46}C_{15}$$

Comparing with $^{99}C_p - {}^{46}C_q$, we get

$$p = 68, \quad q = 15$$

Hence,

$$p + q = 83$$

Therefore, the correct option is D.

Given:

$$x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \cdots + x^{54}(1+x)^{46}$$

Find: A possible value of $p+q$.

From the simplified form used in the solution,

$$S = x^2(1+x)^{99} - x^{55}(1+x)^{46}$$

For $x^2(1+x)^{99}$, the coefficient of $x^{70}$ comes from the coefficient of $x^{68}$ in $(1+x)^{99}$, which is

$${}^{99}C_{68}$$

For $x^{55}(1+x)^{46}$, the coefficient of $x^{70}$ comes from the coefficient of $x^{15}$ in $(1+x)^{46}$, which is

$${}^{46}C_{15}$$

Thus,

$$^{99}C_p - {}^{46}C_q = {}^{99}C_{68} - {}^{46}C_{15}$$

So,

$$p+q = 68+15 = 83$$

Therefore, the correct option is D.