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Let R be a commutative ring. If $\forall x\in R,\ x^{n+2}=x$, show that $2(\sum_{k=0}^n 2^k)x=0.$ (Where $mx:=\sum_{k=1}^m x$).

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    Hint: $\left(2x\right)^{n+2} = 2x$, so $2x = \left(2x\right)^{n+2} = 2^{n+2} x^{n+2}= 2^{n+2} x$. Thus, $\left(2^{n+2}-2\right) x = 0$. – 2017-01-11
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    @darijgrinberg: Given the elementary nature of the problem, I believe that the OP is also expected to prove that $(2x)^{n+2} = 2^{n+2} x^{n+2}$, which is easy, but makes the proof longer than what you wrote in your comment. – 2017-01-11

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First, let's show by induction that if $q \in \Bbb Z$, $k \in \Bbb N$ and $x \in R$, then $q^k x^k = (qx)^k$. If $k=0$, things are clear. The induction step:

$$q^{k+1} x^{k+1} = \sum _{i=0} ^{q^{k+1}} x^{k+1} = \underbrace{ \sum _{i=0} ^{q^k} x^{k+1} + \dots + \sum _{i=0} ^{q^k} x^{k+1} }_{q \text{ times}} = q \sum _{i=0} ^{q^k} x^{k+1} = (qx) \sum _{i=0} ^{q^k} x^k = (qx) (q^k x^k) = \\ (qx) (qx)^k = (qx)^{k+1} .$$

In general, if $q \in \Bbb Z$ and there exist $q' \in \Bbb Z$ such that $q'(q-1)x = x$ for all $x \in R$, then

$$\sum _{k=0} ^p q^k x = q' (q^{p+1} - 1) x .$$

In our case, $p=n$ and $q=2$, so $q' = 1$. Hence,

$$2 \sum \limits _{k=0} ^n 2^k x = 2 (2^{n+1} - 1)x = (2^{n+2} - 2)x = 2^{n+2}x - 2x = 2^{n+2}x^{n+2} - 2x = \\ (2x)^{n+2} -2x = 2x-2x = 0 .$$