Mathematics - Binomial Theorem Question with Solution | TestHub

Given function is $$\begin{array}{r}f(y)=1-(y-1)+(y-1{)}^2-(y-1{)}^3\\ +\dots \dots \dots -(y-1{)}^{17}\end{array}$$ In the expansion of $(y-1{)}^n$ $$T_{r+1}={}^n{\mathrm{C}}_r{y}^{n-r}(-1{)}^r$$ coeff of $y^2$ in $(y-1{)}^2={}^2{\mathrm{C}}_0$ coeff of $y^2$ in $(y-1{)}^3={}^3{\mathrm{C}}_1$ coeff of $y^2$ in $(y-1{)}^4={}^4{\mathrm{C}}_2$ So, coeff of termwise is $$\begin{array}{rl}& {}^2{\mathrm{C}}_0+{}^3{\mathrm{C}}_1+{}^4{\mathrm{C}}_2+{}^5{\mathrm{C}}_3+\dots \dots \dots +{}^{17}{\mathrm{C}}_{15}\\ & =1+{}^3{\mathrm{C}}_1+{}^4{\mathrm{C}}_2+{}^5{\mathrm{C}}_3+\dots \dots \dots ..+{}^{17}{\mathrm{C}}_{15}\\ & =\left({}^3{\mathrm{C}}_0+{}^3{\mathrm{C}}_1\right)+{}^4{\mathrm{C}}_2+{}^5{\mathrm{C}}_3+\dots \dots \dots ..+{}^{17}{\mathrm{C}}_{15}\\ & ={}^4{\mathrm{C}}_1+{}^4{\mathrm{C}}_2+{}^5{\mathrm{C}}_3+\dots \dots \dots +{}^{17}{\mathrm{C}}_{15}\end{array}$$ $$\begin{array}{rl}& ={}^5{\mathrm{C}}_2+{}^5{\mathrm{C}}_3+\dots \dots \dots +{}^{17}{\mathrm{C}}_{15}\\ & ={}^{18}{\mathrm{C}}_{15}={}^{18}{\mathrm{C}}_3\end{array}$$