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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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:

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