Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& x^2(1+x{)}^{98}+x^3\left(1+x^{97}\right)+x^4(1+x{)}^{96}+\dots \dots \\ & x^{54}(1+x{)}^{46}\end{array}$
Coeff. of ${\mathrm{x}}^{70}:{}^{98}{\mathrm{C}}_{68}+{}^{97}{\mathrm{C}}_{67}+{}^{96}{\mathrm{C}}_{66}+\dots \dots \dots$ $\begin{array}{rl}& {}^{47}{\mathrm{C}}_{17}+{}^{46}{\mathrm{C}}_{16}\\ & ={}^{46}{\mathrm{C}}_{30}+{}^{47}{\mathrm{C}}_{30}+\dots \dots \dots ..{}^{98}{\mathrm{C}}_{30}\\ & =\left({}^{46}{\mathrm{C}}_{31}+{}^{46}{\mathrm{C}}_{30}\right)+{}^{47}{\mathrm{C}}_{30}+\dots \dots \dots .{}^{98}{\mathrm{C}}_{30}-{}^{46}{\mathrm{C}}_{31}\\ & ={}^{47}{\mathrm{C}}_{31}+{}^{47}{\mathrm{C}}_{30}+\dots \dots \dots ..{}^{98}{\mathrm{C}}_{30}-{}^{46}{\mathrm{C}}_{31}\\ & \dots \dots \\ & ={}^{99}{\mathrm{C}}_{31}-{}^{46}{\mathrm{C}}_{31}={}^{99}{\mathrm{C}}_{\mathrm{p}}-{}^{46}{\mathrm{C}}_{\mathrm{q}}\end{array}$
Possible values of $(p+q)$ are $62,83,99,46$ $\Rightarrow p+q=83$