Question

(16 choose 0) + (16 choose 1) + ..... + (16 choose 16)
Please Help!

Answer 1

Use the binomial theorem:

`(1+1)^(16)=\displaystyle\sum_(k=0)^(16)\binom{16}k1^(16-k)1^k`

So

`\dbinom{16}0+\dbinom{16}1+\cdots+\dbinom{16}{16}=\boxed{2^(16)}`

More generally,

`\displaystyle\sum_(k=0)^n\binom nk=2^n`