Combinatorial Argument (in Summation Notation)

Question

$$\sum_{k = 0}^{n} {n\choose k} (-2)^{k + 1} $$

I am suffering to answer this question..

Answer 1

I guess that your question is whether there is a simple reduction the the expression you wrote. If so, then the answer is simple:

$$ \sum^n_{k=0}\binom{n}{k}(-2)^{k+1}=-2\sum^n_{k=0}\binom{n}{k}(1)^{n-k}(-2)^k=-2(1-2)^n=-2(-1)^n=2(-1)^{n+1}$$