JEE Main 2024 (Online) 9th April Morning ShiftQuestion 1
The coefficient of $x^{70}$ in $x^{2}(1+x)^{98}+x^{3}(1+x)^{97}+x^{4}(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$ is ${}^{99}C_p-{}^{46}C_q$. Then a possible value of $p+q$ is :
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. The terms form a G.P. with ratio $\frac{x}{1+x}$; after summing, only two binomials are left. Formulas used. (1) G.P. sum: $a+ar+\cdots+ar^{m-1}=a\,\dfrac{1-r^m}{1-r}$(2) ${}^{n}C_r={}^{n}C_{n-r}$Step 1 - The G.P. The series is $\sum_{k=2}^{54}x^k(1+x)^{100-k}$: first term $x^2(1+x)^{98}$, ratio $r=\frac{x}{1+x}$, $53$ terms. Since $1-r=\frac{1}{1+x}$, by (1) $$\text{sum}=x^2(1+x)^{99}\left[1-\frac{x^{53}}{(1+x)^{53}}\right]=x^2(1+x)^{99}-x^{55}(1+x)^{46}.$$ Step 2 - Coefficient of $x^{70}$. From $x^2(1+x)^{99}$ we need $x^{68}$; from $x^{55}(1+x)^{46}$ we need $x^{15}$: $$\text{coefficient}={}^{99}C_{68}-{}^{46}C_{15}.$$ Step 3 - Possible $p,q$. By (2), $p=68$ or $31$ and $q=15$ or $31$. So $p+q\in\{83,\,99,\,46,\,62\}$. Only $83$ is among the options. Answer: (B)