# How to prove $1+1 + 2+ 2^2 +2^3 + \cdots + 2^{30}= 2^{31}$? [closed]

This is true? $1+1 + 2+ 2^2 +2^3 + \cdots + 2^{30}= 2^{31}$, How to prove?

• Do you know anything about geometric series?
– user296602
Commented Feb 13, 2016 at 18:44
• no i dont any thing Commented Feb 13, 2016 at 18:45
• Well, then get a calculator....
– user296602
Commented Feb 13, 2016 at 18:45
• @MichaelHardy I think an elementary-algebra tag is way too vague, especially since this question fits squarely in the scope of algebra-precalculus.
– user296602
Commented Feb 13, 2016 at 19:07
• @MichaelHardy So in your opinion, the fight for changing what "algebra precalculus" means starts with polluting math.SE tags? Commented Feb 14, 2016 at 8:09

First write:

$$\sum_{k=0}^{n+1} x^k = 1 + x\sum_{k=0}^{n} x^k \\ \Rightarrow (1-x)\sum_{k=0}^{n}x^k = 1 - x^{n+1} \\ \Rightarrow \sum_{k=0}^{n}x^k = \frac{1 - x^{n+1}}{1 - x}$$

$$1 + \sum_{k=0}^{30} 2^k = 1 + \frac{1 - 2^{31}}{1- 2}\\ = 1 + 2^{31} - 1 \\ = 2^{31}$$
If you don't know anything about geometric series, $2^n-1=1+2+\cdots+2^{n-1}$ can be easily proved by induction.
It's clearly true for $n=1$. Suppose it's true for some $n=k$ and consider $n=k+1$.
$$1+2+\cdots+2^{k-1}+2^k=(2^k-1)+2^k=2\cdot 2^k-1=2^{k+1}-1.$$
For computing the geometric sum $S=1+x+x^2+\cdots+x^n$ compare it to $$x\cdot S=x+x^2+x^3+\cdots+x^{n+1}$$ to find that the difference is $$(x-1)\cdot S=x^{n+1}-1$$ Now set $x=2$…