JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
Suppose $2-p,\ p,\ 2-\alpha,\ \alpha$ are the coefficients of four consecutive terms in the expansion of $(1+x)^{n}$. Then the value of $p^{2}-\alpha^{2}+6\alpha+2p$ equals
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. The four coefficients pair up: $(2-p)+p=2$ and $(2-\alpha)+\alpha=2$. Pascal's rule turns each pair into one coefficient of $(1+x)^{n+1}$. Formulas used. (1) Pascal's rule: ${}^{n}C_r+{}^{n}C_{r+1}={}^{n+1}C_{r+1}$(2) ${}^{N}C_j={}^{N}C_k$ with $j\ne k$ $\Rightarrow$ $j+k=N$Step 1 - Pair the coefficients. Let them be ${}^{n}C_r,{}^{n}C_{r+1},{}^{n}C_{r+2},{}^{n}C_{r+3}$. By (1), $${}^{n+1}C_{r+1}=(2-p)+p=2,\qquad{}^{n+1}C_{r+3}=(2-\alpha)+\alpha=2.$$ Step 2 - Find $n$. By (2), $(r+1)+(r+3)=n+1$, so $n=2r+3$. Step 3 - The middle two are equal. $(r+1)+(r+2)=2r+3=n$, so ${}^{n}C_{r+1}={}^{n}C_{r+2}$, i.e. $p=2-\alpha$ and $p+\alpha=2$. Step 4 - Value. $$p^2-\alpha^2+6\alpha+2p=(p-\alpha)(p+\alpha)+6\alpha+2p=2(p-\alpha)+6\alpha+2p=4(p+\alpha)=8.$$ Answer: (C)