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Exercise 414Combinatorics › Binomial coefficientsHigh schoolAdded 3 Nov 2026

Sum of powers of 2

#power sum#2^n#binomial

Statement

Compute ∑k=0n2k(nk)\sum_{k=0}^{n}2^k\binom{n}{k}.

Hint

Set a=1a=1, b=2b=2 in the binomial theorem.

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Solution

Binomial theorem. (1+2)n=∑k(nk)2k(1+2)^n=\sum_k\binom{n}{k}2^k.

Conclusion. ∑k2k(nk)=3n\sum_k 2^k\binom{n}{k}=3^n.