Question

C1 + C2 + C3 + ..... C8​

Answer 1

Answer:

If you are talking about binomial

$(1 + x) {}^{8} = \sum _{k = 0}^{8} \: \binom{8}{k}.{ {x}^{k} }$

put x = 1

$\sum _{k = 0}^{8} \: \binom{8}{k} = {2}^{8}$