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I want to prove that for all positive integers $n$, $2^1+2^2+2^3+...+2^n=2^{n+1}-2$. By mathematical induction:

1) it holds for $n=1$, since $2^1=2^2-2=4-2=2$

2) if $2+2^2+2^3+...+2^n=2^{n+1}-2$, then prove that $2+2^2+2^3+...+2^n+2^{n+1}=2^{n+2}-2$ holds.

I have no idea how to proceed with step 2), could you give me a hint?

Man_Of_Wisdom
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anoldguy
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    For (2): add $2^{n+1}$ on both sides. – AdLibitum Apr 19 '15 at 10:21
  • Another possibility is that you use the formula for the geometric series (finite series): $\sum_{k=1}^{n}2^k=\sum_{k=0}^{n}2^k-1=\frac{1-2^{n+1}}{1-2}-1=2^{n+1}-2$ – Marm Apr 19 '15 at 15:05

3 Answers3

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It is straightforward

$$2+2^2+\cdots+2^n+2^{n+1}=(2+2^2+\cdots+2^n)+2^{n+1}$$

Use the induction assumption

$$\begin{align} 2+2^2+\cdots+2^n+2^{n+1}&=2^{n+1}-2+2^{n+1}\\&=2\cdot 2^{n+1}-2\\&=2^{n+2}-2\end{align}$$

marwalix
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start with zero, the question is then the same as $$2^0+2^1+2^2+...+2^n=2^{n+1}-1$$ The binary for the LHS is $$ 11111\dots 1 $$ with $n$ ones. add $1$ to get $$ 10000\dots 0, $$ with $n$ zeros in binary. This is the binary for $2^{n+1}$ and are done.

Mark Joshi
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Define $S=2^1+\cdots+2^n$ and realize that $2S=2^2+\cdots+2^{n+1}=S+2^{n+1}-2$.

Michael Hoppe
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