Question

The sum of the series $\sum _{r=1}^n(-1{)}^{r-1}{.}^n{C}_r(a-r)$ is

• A. $a$
• B. $0$
• C. $n{.2}^{n-1}+a$
• D. None of these

Answer 1

The correct option is A $a$

$\begin{array}{cc}{\sum }_{r=1}^n{\left(-1\right)}^{r-1}{}^n{c}_r\left(a-r\right)& \\ Putn=1& \\ {\Rightarrow }^n{c}_1\left(a-1\right){-}^n{c}_2\left(a-2\right){+}^n{c}_3\left(a-1\right)+.......{\left(-1\right)}^{n-1}{}^n{c}_n\left(a-n\right)& \\ \Rightarrow a\left[{}^n{c}_1{-}^n{c}_2{+}^n{c}_3{-}^n{c}_4{+}^n{c}_5+.......{-}^n{c}_n\right]{-}^n{c}_1+2^n{c}_2-3^n{c}_3+4^n{c}_4.........& \\ =a\left[{}^n{c}_1{-}^n{c}_2{+}^n{c}_3{-}^n{c}_4{+}^n{c}_5+......\right]-\left[{}^n{c}_1-2^n{c}_2+3^n{c}_3+4^n{c}_4+.......\right]& \\ {\left(1-x\right)}^n{=}^n{c}_0{-}^n{c}_{1}x{+}^n{c}_2{x}^2{-}^n{c}_3{x}^3.......& \left[Accordingtothisformula\right]\\ \Rightarrow Add\&subtract{}^n{c}_{0}inahaveequation\&takingoutcommon\left(-1\right)also.& \\ \Rightarrow a\left[-\left({}^n{c}_0{-}^n{c}_1{+}^n{c}_2{-}^n{c}_3+......\right){+}^n{c}_o\right]-\left[{}^n{c}_1-2^n{c}_2+3^n{c}_3-4^n{c}_4\right]& \\ Putx=1& \\ {\left(1-1\right)}^n{=}^n{c}_0{-}^n{c}_1{+}^n{c}_2{-}^n{c}_3=0& \\ So,& \\ a\left[-0{+}^n{c}_0\right]-\left[{}^n{c}_1-2^n{c}_2+3^n{c}_3-4^n{c}_4+.........\right]\to \left(i\right)& \\ {\left(1+n\right)}^n{=}^n{c}_0{+}^n{c}_{1}x{+}^n{c}_2{x}^2{+}^n{c}_3{x}^3{+}^n{c}_4{x}^4+.......& \left[\leftarrow formula\right]\\ Diff.w.r.t.n,weget& \\ n{\left(1+x\right)}^{n-1}{=}^n{c}_1+2x^n{c}_2+3x^n{c}_3+4x^n{c}_4+....& \\ Putx=-1So,& \\ x{\left(1-x\right)}^{n-1}{=}^n{c}_1-2x^n{c}_2+3{x^2}^n{c}_3+4^n{c}_4+.........& \\ 0{=}^n{c}_1-2^n{c}_2+3^n{c}_3-4^n{c}_4& \end{array}$

So, put the value in equation in (1)

$\begin{array}{c}=a\left[{}^n{c}_0\right]-\left[0\right]\Rightarrow a\left(1\right)=a\\ {\sum }_{r=1}^n{\left(-1\right)}^{r-1}{}^n{c}_r\left(a-r\right)=a\end{array}$