Find value of $\sum\limits_{k=0}^{10}{(-1)^k C(10,k)/(2^k)}$

Question

Find value of $\sum\limits_{k=0}^{10}{(-1)^k\dbinom{10}{k}\dfrac{1}{2^k}}$

Do I have to open the factorials of all combinations, or is there any other way? please help.

Answer 1

You have $$(x-1)^{10} = \sum_{k=1}^{10}C(10,k)x^k(-1)^{10-k}$$.

Note that $(-1)^{10-k} = (-1)^k$, for all $0 \leq k \leq 10$.

Taking $x=\frac{1}{2}$, you have a result.

Answer 2

HINT $$\sum_{k=0}^{10}\left(\frac{-1}{2}\right)^k\cdot \binom{10}{k}\cdot1=\left(1-\frac{1}{2}\right)^{10}$$