nC0 + nC1 + nC2 + ... + nCn = 2^n

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Question from Chinonyerem, a student:

For n >= 1, derive the identity
nC0 + nC1 + nC2 + ... + nCn = 2^n
[Hint: Let a = b = 1 in the binomial theorem]

Chinonyerem,

There is a statement of the binomial theorem in my response to an earlier question. If you follow the hint and let a = b = 1 then the statement of the binomial theorem becomes exactly the expression you are asked to derive.

Penny

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