JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 8)
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 $${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$$. Then a possible value of $$\mathrm{p}+\mathrm{q}$$ is :
61
83
55
68
Explanation
$$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$$
It is a G.P. with first term $$=x^2(1+x)^{98}$$
and common ratio $$=\frac{x}{1+x}$$
sum of these term $$=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$$
$$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$$
_9th_April_Morning_Shift_en_8_1.png)
$$\begin{aligned} & ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\ & \Rightarrow p=68, q=15 \\ & \Rightarrow p+q=83 \end{aligned}$$
Comments (0)
