The coefficient of x^70 in x^2(1+x)^98+x^3(1+x)^97+x^4(1+x)^96+ +x^54(1+x)^46 is ^99C_p-^4…

Question

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 :

Accepted Answer

$$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$$

It is a G.P. with first term $=x^2(1+x)^{98}$

and common ratio $$=\frac{x}{1+x}$$

sum of these term $$=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$$

$$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$$

$$\begin{aligned}
& ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \
& \Rightarrow p=68, q=15 \
& \Rightarrow p+q=83
\end{aligned}$$

Solutions · 1