JEE Main 2026 (Online) 23rd January Morning ShiftQuestion 1
The sum of all possible values of $n\in\mathbf{N}$, so that the coefficients of $x$, $x^2$ and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is :
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Expand $(1+x^2)^2$ and collect the coefficients of $x$, $x^2$, $x^3$. Formulas used. (1) $(1+x^2)^2=1+2x^2+x^4$(2) Three numbers $p,q,s$ are in A.P. iff $2q=p+s$Step 1 - Coefficients. Multiply $1+2x^2+x^4$ by $(1+x)^n=\sum{}^{n}C_rx^r$: $$x:\ n,\qquad x^2:\ {}^{n}C_2+2,\qquad x^3:\ {}^{n}C_3+2n.$$ Step 2 - A.P. condition. By (2), $$2\left({}^{n}C_2+2\right)=n+{}^{n}C_3+2n.$$ Multiply by $6$: $$6n^2-6n+24=n^3-3n^2+20n\ \Longrightarrow\ n^3-9n^2+26n-24=0.$$ Step 3 - Solve. $n=2$ is a root; dividing gives $(n-2)(n^2-7n+12)=(n-2)(n-3)(n-4)=0$. So $n=2,3,4$. Check: $n=2$ gives $2,3,4$; $n=3$ gives $3,5,7$; $n=4$ gives $4,8,12$. All are A.P.s. Step 4 - Sum. $2+3+4=9$. Answer: (C)