# Why is $\sum_{i=0}^{w-1}{2^i} = 2^w - 1$? [duplicate]

The title says it all, I guess. Why $$\sum_{i=0}^{w-1}{2^i} = 2^w - 1$$

## marked as duplicate by JMoravitz, clathratus, cmk, John Omielan, StubbornAtomSep 11 at 18:17

• Call the sum $S(w)$. Can you relate $S(w)$ to $2S(w)$ in some helpful way? – lulu Sep 11 at 17:46
• Do you recognize how/why $999999+1 = 1000000$? If we were in binary instead, why $111111_2+1_2=1000000_2$? As for a proof, have you ever heard of a proof by induction? – JMoravitz Sep 11 at 17:47
• What have you tried? Can you show it is true for some special value of w? Are you familiar with induction? – M_B Sep 11 at 17:47
• Yea I know that I could prove it by induction but isn't there some algebraic reasoning for the formula? – That Guy Sep 11 at 17:50

Here's another way to look at it. \begin{align}S=1+&2+4+8+\cdots+2^n\\ 2S=\ \ &2+4+8+\cdots+2^n+2^{n+1}\end{align} Subtract the first equation from the second to get $$S=2^{n+1}-1$$

Here is an intuitive way to see it:

Say you want to compute $$1+2+2^2+2^3+2^4\tag{1}$$ If you add $$1$$ to the sum $$(1)$$ you'll get: $$(1+1)+2+2^2+2^3+2^4\\=(2+2)+2^2+2^3+2^4\\=(2^2+2^2)+2^3+2^4\\=(2^3+2^3)+2^4\\=(2^4+2^4)\\=2^5$$

Since we added $$1$$, the sum is $$1$$ less, that is, $$2^5-1$$

Say we set $$s_n(x)=\sum_{k=0}^{n-1}x^k.$$ We see that $$s_{n+1}(x)=x^n+s_n(x).$$ But we also see that $$xs_n(x)=\sum_{k=0}^{n-1}x^{k+1}=\sum_{k=1}^{n}x^k=s_{n+1}(x)-1.$$ Hence we have $$xs_n(x)=s_n(x)+x^n-1,$$ which means that for $$x\ne 1$$, $$s_n(x)=\frac{x^n-1}{x-1}.$$ The sum in question is given by $$\sum_{i=0}^{w-1}2^i=s_{w}(2)=\frac{2^w-1}{2-1}=2^w-1.$$