JEE Main 2026 — Binomial Theorem Question with Solution

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

The coefficient of $x^3$ in ${\sum }_{r=3}^{99}(1+x{)}^r+(1+kx{)}^{100}$ is:

${\sum }_{r=3}^{99}{}^r{C}_3+k^3{}^{100}{C}_3$

Using the identity ${\sum }_{r=k}^n{}^r{C}_k={}^{n+1}{C}_{k+1}$, we get:

${\sum }_{r=3}^{99}{}^r{C}_3={}^{100}{C}_4$

Thus, the total coefficient of $x^3$ is ${}^{100}{C}_4+k^3{}^{100}{C}_3$.

Given that this coefficient is equal to $\left(43n+\frac{101}{4}\right){}^{100}{C}_3$, we have:

${}^{100}{C}_4+k^3{}^{100}{C}_3=\left(43n+\frac{101}{4}\right){}^{100}{C}_3$

Dividing both sides by ${}^{100}{C}_3$:

$\frac{{}^{100}{C}_4}{{}^{100}{C}_3}+k^3=43n+\frac{101}{4}$

Using $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$, we get $\frac{{}^{100}{C}_4}{{}^{100}{C}_3}=\frac{100-4+1}{4}=\frac{97}{4}$.

Substituting this value:

$\frac{97}{4}+k^3=43n+\frac{101}{4}$

$k^3=43n+\frac{101}{4}-\frac{97}{4}$

$k^3=43n+1$

$k^3-1=43n$

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

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

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

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

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

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

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

Answer: $11$