Question
The sum of the series $$\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot {\,^n}{C_r}} \left( {a - r} \right)$$ is equal to
A.
$$n \cdot {2^{n - 1}} + a$$
B.
$$0$$
C.
$$a$$
D.
None of these
Answer :
$$a$$
Solution :
Sum $$ = a\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot {\,^n}{C_r}} - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot r \cdot {\,^n}{C_r}} $$
$$\eqalign{
&\,\,\,\,\,\,\,\,\,\,\,\,\, = a\left\{ {^n{C_1} - {\,^n}{C_2} + {\,^n}{C_3} - .....} \right\} - n\left\{ {^{n - 1}{C_0} - {\,^{n - 1}}{C_1} + .....} \right\}\left( {{\text{because }}r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}} \right) \cr
&\,\,\,\,\,\,\,\,\,\,\,\,\, = a \cdot {\,^n}{C_0}\,\,\left( {\because \,{\,^n}{C_0} - {\,^n}{C_1} + {\,^n}{C_2} - ..... = 0} \right). \cr} $$