Mathematics - Binomial Theorem Question with Solution | TestHub

$T_{k+1}=(-1{)}^k\cdot {}^n{C}_k(a-k)(b-k)$ $=(-1{)}^k\cdot {}^n{C}_k\left[ab-k(a+b)+k^2\right]$

Thus, the sum of series in (i) $\begin{array}{rlr}& =\sum _{k=0}^n(-1{)}^k\cdot {}^n{C}_k\left[ab-k(a+b)+k^2\right]& =ab\sum _{k=0}^n(-1{)}^k\cdot {}^n{C}_k-(a+b)\sum _{k=0}^n(-1{)}^{k}k\cdot {}^n{C}_k+\sum _{k=0}^n(-1{)}^k{k}^2\cdot {}^n{C}_k\end{array}$ We know that $(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\dots .+{}^n{C}_n{x}^n\dots \text{ (ii) }$ Differentiating both sides w.r.t. x, we get $n(1+x{)}^{n-1}=1\cdot {}^n{C}_1+2\cdot {}^n{C}_{2}x+3\cdot {}^n{C}_3{x}^2+\dots .+n\cdot {}^n{C}_n{x}^{n-1}\dots \text{.(iii) }$ Multiplying it by x we get $n\cdot x(1+x{)}^{n-1}=1\cdot {}^n{C}_{1}x+2\cdot {}^n{C}_2{x}^2+3\cdot {}^n{C}_3{x}^3+\dots .+n\cdot {}^n{C}_n{x}_n$ Differentiating w.r.t. x, we get $\begin{array}{rlr}& n(1+x{)}^{n-1}+n(n-1)x(1+x{)}^{n-2}& =1^2\cdot {}^n{C}_1+2^2\cdot {}^n{C}_{2}x+\dots .+n^2\cdot {}^n{C}_n{x}^{n-1}\dots \text{ (iv) }\end{array}$ Putting $\mathrm{x}=-1$ in (ii), (iii) and (iv), we get $\begin{array}{rlrl}& \sum _{k=0}^n(-1{)}^k\cdot {}^n{C}_k=0& \sum _{k=0}^n(-1{)}^{k}k\cdot {}^n{C}_k=0& \sum _{k=0}^n(-1{)}^k{k}^2{}^n{C}_k=0\end{array}$

Thus, the sum of the series is 0