Mathematics - Binomial Theorem Question with Solution | TestHub

We have $(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\dots \dots$.

$$\begin{array}{rlrlr}& +{}^n{C}_{r-1}{x}^{r-1}+{}^n{C}_r{x}^r+\dots \dots +{}^n{C}_n{x}^n\dots .& \text{ and }(1+x{)}^{-3}=1-{}^3{C}_{1}x+{}^4{C}_2{x}^2\dots \dots +& (-1{)}^{r-1}{}^{r+1}{C}_{r-1}{x}^{r-1}+(-1{)}^r& {}^{r+2}{C}_r{x}^r+\dots \dots ..to\infty \end{array}$$

Multiply (1) and (2), we get $(1+x{)}^{n-3}=\left({}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\dots \dots +{}^n{C}_n{x}^n\right)\left(1-{}^3{C}_{1}x+{}^4{C}_2{x}^2-\dots .\infty \right)$ Clearly, the coefficient of $x^r$ from the product in R.H.S is $1.{}^n{C}_r-{}^3{C}_1.{}^n{C}_{r-1}+{}^4{C}_2.{}^n{C}_{r-2}-\dots \dots +(-1{)}^r{}^{r+2}{C}_r.{}^n{C}_0$ $={}^n{C}_r-3{}^n{C}_{r-1}+\frac{4.3}{2.1}{}^n{C}_{r-2}-\frac{5.4}{2.1}{}^n{C}_{r-3}$ $+\dots \dots +(1{)}^r\frac{(r+2)(r+1)}{2.1}$ $=\frac{1}{2}\left[1.2{}^n{C}_r-2.3{}^n{C}_{r-1}+4.3{}^n{C}_{r-2}-\dots \dots +(-1)r(r+2)(r+1)\right]$ ∴ Required series $=2\times$ coefficient of $x^r$ in $(1+x{)}^{n-3}=2.{}^{n-3}{C}_r$