1. Apr 26, 2006

### nick727kcin

how does this come out to this?

i dont understand

2. Apr 26, 2006

### siddharth

What part do you not understand?

3. Apr 26, 2006

### Curious3141

What part don't you get ? It's just algebraic rearrangement :

$$2^{2n+1}5^{-n} = 2^{2n}(2)(5^{-n}) = (2)({(2^2)}^n)(5^{-n}) = (2)(4^n)(\frac{1}{5^n}) = (2)(\frac{4^n}{5^n}) = (2){(\frac{4}{5})}^n$$

I would've just left it there since it's obvious this is a geometric series, but the text apparently felt it necessary to take another step (which seems rather pointless), but it's also just algebra.

$$(2){(\frac{4}{5})}^n = (2)(\frac{4}{5}){(\frac{4}{5})}^{n-1} = (\frac{8}{5}){(\frac{4}{5})}^{n-1}$$

Clear enough ?

4. Apr 26, 2006

### HallsofIvy

Staff Emeritus
22n+1=(2)22n= (2)(22)n= (2)(4n).
5-n= 1/5n

so
22n+15-n= (2)(4n)/5n= (2)(4/5)n.

Curious3141, the reason for "next step" is that the sum starts at n= 1. The standard formula for the sum of a geometric series,
$$\Sigma_{n=0}^\infty ar^n= \frac{a}{1-r}$$
requires that the sum start at n=0.
$$\Sigma_{n=1}^\infty$$\frac{8}{5}$$$$\frac{4}{5}$$^{n-1}= \Sigma_{n=0}^\infty$$\frac{8}{5}$$$${4}{5}$$^n$$

Last edited: Apr 26, 2006