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I should simplify the following expression (for a complex number): $$\sum _{ k=0 }^{ n }{ \binom{n}{k}}i^{k}3^{k-n} $$

The solution is $(i+\frac{1}{3})^n$,but i don't quite get the steps. If would be nice if someone could explain.


The Binomial Theorem: $(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k}$

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$3^{k-n}=\left(\dfrac{1}{3}\right)^{n-k}$ –  M. Strochyk Mar 21 '13 at 21:52

3 Answers 3

up vote 4 down vote accepted

$$ \sum_{k=0}^n \binom{n}{k}i^k 3^{k-n} = \sum_{k=0}^n \binom{n}{k}i^k \left(3^{-1}\right)^{n-k} = \sum_{k=0}^n \binom{n}{k}i^k \left(1/3\right)^{n-k} = (i+1/3)^n $$

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Just use that $$3^{k-n}=\left(\frac{1}{3}\right)^{n-k}$$ Then you have exactly the binomial theorem and hence it is $$(i+\frac{1}{3})^n$$

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Hint: Set $x = \frac {1}{3}$, $ y = i$.

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thats the solution which is given, how should that help the OP ? –  Dominic Michaelis Mar 21 '13 at 22:00

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