Question

How do you find the formula for the sum of this?

`sum_(r=1)^(n)2^r`

Answer 1

`sum_(r=1)^n 2^r=color(blue)(2(2^n-1))`

Explanation:

`sum_(r=1)^n 2^r=2^1+2^2+2^3+...+2^n`

`color(white)("XXX")=2+2(2^1+2^2+...+2^(n-1))`

`color(white)("XXX")=2+2(sum_(r=1)^n 2^r-2^n)`

`color(white)("XXX")=2+2sum_(r=1)^n 2^r-2^(n+1)`

Subtracting `sum_(r=1)^n 2^r` from both sides
`0=2+sum_(r=1)^n 2^r-2^(n+1)`

then adding `2^(n+1)-2` to both sides
`2^(n+1)-2=sum_(r=1)^n 2^r`

switching sides
`sum_(r=1)^n 2^r=2^(n+1)-2`
`color(white)("XXX")=2(2^n-1)`

Answer 2

`sum_(r=1)^n 2^r = 2(2^n-1)`

Explanation:

If we expand the first few terms of the sum we have:

`sum_(r=1)^n 2^r = 2^1+2^2+2^3+ ... + 2^n`

And we notice the terms form a Geometric Progression with first term `a=2` and common ratio `r=2` also. Thuis using the GP summation formula:

`S_n = a(1-r^n)/(1-r)`

We have:

`sum_(r=1)^n 2^r = (2){ (1-2^n)/(1-2)}`
`\ \ \ \ \ \ \ \ \ \ = 2{ (1-2^n)/(-1)}`
`\ \ \ \ \ \ \ \ \ \ = 2(2^n-1)`