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 99Cp – 46Cq.

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 : 

(1) 55 

(2) 61 

(3) 68 

(4) 83

Answer 1

Correct option is : (4) 83 

\( \text{S}=x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots .+x^{54}(1+x)^{46} \) 

It is a G.P. 

\( \text{S}=x^2(1+x)^{98}\left[\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right] \) 

Coefficient of  \(x^{70}\) in \(\text{S}=x^2(1+x)^{46}\left[(1+x)^{53}-x^{53}\right]\) 

Coefficient of \(x^{70}\) in \(\text{S}=x^2(1+x)^{99}-x^{55}(1+x)^{46}\) 

\( \begin{aligned} \mathrm{S}={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}} \end{aligned} \) 

\( \begin{aligned} & \mathrm{P}=68, \ \mathrm{q}=15 \\ \end{aligned} \)

\( \begin{aligned} & \mathrm{P}+\mathrm{q}=83 \end{aligned} \)