1. ## Series question

Hi
The question given is

Prove by mathematical induction that

$\displaystyle \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{n}{3^n} = \frac{1}{4}(3-\frac{2n + 3}{3^n})$

I've spent quite a bit of time on this, playing around with it, but can't seem to make any head-way. If anyone can give any hints or tips at all i'd be rather grateful.

-Dave

2. Originally Posted by walleye
Hi
The question given is

Prove by mathematical induction that

$\displaystyle \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{n}{3^n} = \frac{1}{4}(3-\frac{2n + 3}{3^n})$

I've spent quite a bit of time on this, playing around with it, but can't seem to make any head-way. If anyone can give any hints or tips at all i'd be rather grateful.

-Dave

For n = 1 we get:

Left side: $\displaystyle \frac{1}{3}$

Right side:$\displaystyle \frac{1}{4}\left(3-\frac{2\cdot 1+3}{3^1}\right)=\frac{1}{4}\left(3-\frac{5}{3}\right)=\frac{1}{4}\left(\frac{4}{3}\ri ght)=\frac{1}{3}$...check.

Assume now the claim's true for n and let's try to show for n+1:

$\displaystyle \frac{1}{3}+\frac{2}{3^2}+\ldots+\frac{n}{3^n}+\fr ac{n+1}{3^{n+1}}=\frac{1}{4}\left(3-\frac{2n+3}{3^n}\right)+\frac{n+1}{3^{n+1}}$ $\displaystyle =\frac{1}{4}\left(\frac{3^{n+1}-3-2n}{3^n}+\frac{4n+4}{3^{n+1}}\right)$ $\displaystyle = \frac{1}{4}\left(\frac{3^{n+2}-9-6n+4n+4}{3^{n+1}}\right)$ .

Well, now with a little (very little, indeed) algebra check that the rightmost side equals the formula with n+1 instead n, and we're done.

Tonio

3. I'm sorry, I forgot to thank you for this.

after a bit of reworking I finally got it.

Thanks

4. Tonio, sorry, doesn't your last expression evaluate to $\displaystyle \frac{1}{4}\left(3-\frac{2n+\color{red}5}{3^{n+1}}\right)$ ?

5. Originally Posted by scorpion007
Tonio, sorry, doesn't your last expression evaluate to $\displaystyle \frac{1}{4}\left(3-\frac{2n+\color{red}5}{3^{n+1}}\right)$ ?

Indeed it does...just as needed.

Tonio

6. But don't we need

$\displaystyle \frac{1}{4}\left(3-\frac{2n+3}{3^{n+1}}\right)$
?

Ah, nevermind! I forgot to substitute n+1 in 2(n+1)

7. You can also differentiate $\displaystyle 1+x+x^2+\dots+x^n = \frac{1-x^{n+1}}{1-x}$ with respect to $\displaystyle x$, multiply both sides by $\displaystyle x$ and set $\displaystyle x=\frac{1}{3}$.