SolveItJEE Main
Mathematics · 2026

JEE Main · 4 April 2026, Shift 1 · Q9

Let the smallest value of k ∈ N, for which the coefficient of x^3 in (1+x)^3+(1+x)^4+(1+x)^5+…+(1+x)^99+(1+k x)^100, x ≠ 0, is (43 n+101/4)(^100 C_3)…

Let the smallest value of $\displaystyle k \in \mathrm{~N}$, for which the coefficient of $\displaystyle x^3$ in $\displaystyle (1+x)^3+(1+x)^4+(1+x)^5+\ldots+(1+x)^{99}+(1+k x)^{100}, x \neq 0$, is $\displaystyle \left(43 n+\frac{101}{4}\right)\left({ }^{100} \mathrm{C}_3\right)$ for some $\displaystyle n \in \mathrm{~N}$, be $\displaystyle p$. Then the value of $\displaystyle p+n$ is:
ShareWhatsAppTelegram

More from Binomial Theorem

JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.