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

User Vaughan
by
8.0k points

1 Answer

3 votes

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

User Laura Calinoiu
by
8.3k points

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