$\sum_{i=-1}^{99} 2^i$ Solution Assistance

Question

So i have a general understanding of summations

The problem i have is this $$\sum_{i=-1}^{99}2^i$$

I simplified it to $$\sum_{i=1}^{101} 2^{i - 2}$$ But i keep getting stuck here when trying to eliminate the sum by subbing in for i.

What do you want to do with that sum other than rewrite it? If you want to evaluate it, write out the first few terms to see the pattern. Starting from either form should give the same first few terms. Then see if you can compare the sum to the sum $1 + 2 + 4 + \text{ a few more terms}$, which is not hard to sum. — Ethan Bolker, Mar 12 '18 at 20:20

score 2 · Answer 1 · answered Mar 12 '18 at 20:19

2

Just write $2^{-1}+\sum_{i=0}^{99}2^i$ and use $$\sum_{k=0}^n2^k=2^{n+1}-1.$$

answered Mar 12 '18 at 20:19

Dietrich Burde

130,978

score 2 · Answer 2 · answered Mar 12 '18 at 20:23

$$2^{-1}+2^{0}+2^{1}+\dots+2^{99}=:S$$ Now calculate $2S$: $$2S=2(2^{-1}+2^{0}+2^{1}+\dots+2^{99})$$ $$2S=2^0+2^1+\dots+2^{100}$$ Now substract them: $$2S-S=(2^0+2^1+\dots+2^{100})-(2^{-1}+2^{0}+2^{1}+\dots+2^{99})$$ $$S=(2^{100}+2^{99}+2^{98}+\dots+2^1+2^0)-(2^{99}+2^{98}+2^{1}+\dots+2^0+2^{-1})$$ $$S=2^{100}-2^{-1}$$ $$S=\frac{2^{101}-1}{2}$$

score 0 · Answer 3 · answered Mar 12 '18 at 20:23

0

$$\sum_{i=-1}^{99}2^i=\frac12+\sum_{i=0}^{99}2^i=\frac12+\frac{2^{99+1}-1}{2-1}=2^{100}-\frac12$$

answered Mar 12 '18 at 20:23

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

score 0 · Answer 4 · answered Mar 12 '18 at 20:29

Let's look at this sum, we can recognize that it is a geometric sum:

$$\sum_{i=-1}^{99}2^i$$

The first term, $a$, is obtained by substituting the first index: $$a=2^{-1}$$

The common ratio, $r=2$.

Also, let's get the number of terms, $n=99-(-1)+1=101$.

Now, let's use the formula:

$$\frac{a(r^n-1)}{r-1}=\frac{2^{-1}(2^{101}-1)}{2-1}=2^{-1}(2^{101}-1)=2^{100}-2^{-1}$$

$\sum_{i=-1}^{99} 2^i$ Solution Assistance

4 Answers4