I need the proof for the following equation/theorem:
$$n^k = \sum_{i=0}^{n-1}\sum_{j=1}^{k}\binom{k}{j}i^{k-j}$$
How do I prove that $n^k$ is equal to this summation-of-a-summation: $n^k = \sum_{i=0}^{n-1}\sum_{j=1}^{k}\binom{k}{j}i^{k-j}$?
-5
$\begingroup$
summation
binomial-coefficients
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0What did you try ? – 2017-01-26
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1@YvesDaoust My guess is that we will never know since you simultaneously posted your answer. Is the comment supposed to justify answering this question with no context? – 2017-01-26
1 Answers
1
By the Binomial theorem
$$\sum_{j=1}^{k}\binom{k}{j}i^{k-j}=(1+i)^k-i^k$$ and the claim follows by telescoping.