JEE Main 2026 — Mathematics Algebra

The coefficient of $x^3$x3 in the given expression is the sum of the coefficients of $x^3$x3 in each term.

The coefficient of $x^3$x3 in $\displaystyle \sum_{r=3}^{99} (1+x)^r + (1+kx)^{100}$r=3∑99​(1+x)r+(1+kx)100 is:

$$\sum_{r=3}^{99} {}^{r}C_3 + k^3 {}^{100}C_3$$r=3∑99​rC3​+k3100C3​

Using the identity $\displaystyle \sum_{r=k}^{n} {}^{r}C_k = {}^{n+1}C_{k+1}$r=k∑n​rCk​=n+1Ck+1​, we get:

$$\sum_{r=3}^{99} {}^{r}C_3 = {}^{100}C_4$$r=3∑99​rC3​=100C4​

Thus, the total coefficient of $x^3$x3 is ${}^{100}C_4 + k^3 {}^{100}C_3$100C4​+k3100C3​.

Given that this coefficient is equal to $\left(43n + \dfrac{101}{4}\right) {}^{100}C_3$(43n+4101​)100C3​, we have:

$${}^{100}C_4 + k^3 {}^{100}C_3 = \left(43n + \dfrac{101}{4}\right) {}^{100}C_3$$100C4​+k3100C3​=(43n+4101​)100C3​

Dividing both sides by ${}^{100}C_3$100C3​:

$$\dfrac{^{100}C_4}{^{100}C_3} + k^3 = 43n + \dfrac{101}{4}$$100C3​100C4​​+k3=43n+4101​

Using $\dfrac{^{n}C_r}{^{n}C_{r-1}} = \dfrac{n-r+1}{r}$nCr−1​nCr​​=rn−r+1​, we get $\dfrac{^{100}C_4}{^{100}C_3} = \dfrac{100-4+1}{4} = \dfrac{97}{4}$100C3​100C4​​=4100−4+1​=497​.

Substituting this value:

$$\dfrac{97}{4} + k^3 = 43n + \dfrac{101}{4}$$497​+k3=43n+4101​

$$k^3 = 43n + \dfrac{101}{4} - \dfrac{97}{4}$$k3=43n+4101​−497​

$$k^3 = 43n + 1$$k3=43n+1

$$k^3 - 1 = 43n$$k3−1=43n

Since $n \in \mathbb{N}$n∈N, we must have $n \ge 1$n≥1, which implies $k^3 - 1 \ge 43 \Rightarrow k \ge 4$k3−1≥43⇒k≥4.

Checking the values of $k \ge 4$k≥4 to find the smallest integer $k$k such that $k^3 - 1$k3−1 is divisible by $43$43:

For $k = 4$k=4, $k^3 - 1 = 63$k3−1=63 (not divisible by $43$43).

For $k = 5$k=5, $k^3 - 1 = 124$k3−1=124 (not divisible by $43$43).

For $k = 6$k=6, $k^3 - 1 = 215 = 43 \times 5$k3−1=215=43×5.

Thus, the smallest value of $k$k is $p = 6$p=6, and the corresponding value of $n$n is $5$5.

Therefore, $p + n = 6 + 5 = 11$p+n=6+5=11.

Answer: $11$11