# Math Help - Stuck on an Induction Problem

1. ## Stuck on an Induction Problem

Prove that:
$\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{n}{3^n} = \frac{3}{4} - \frac{2n+3}{4*3^n}$

My work so far:
Let P(n) denote the statement that $\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{n}{3^n} = \frac{3}{4} - \frac{2n+3}{4*3^n}$

$P(1) = \frac{1}{3^1} = \frac{3}{4} - \frac{5}{4*3^1}$
$\frac{1}{3} = \frac{1}{3}$
The statement holds true for the case n = 1.

Assume that P(n) is true for when n is equal to a natural constant k,
$P(k) = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{k}{3^k} = \frac{3}{4} - \frac{2k+3}{4*3^k}$

Prove that P(n) is true when n = k + 1
$P(k+1) = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + ... + \frac{k}{3^k} + \frac{k+1}{3^{k+1}}= \frac{3}{4} - \frac{2(k+1)+3}{4*3^{k+1}}$

Replacing the terms on the left hand side of P(k+1) with the right hand side of P(k) I then had:

$P(k+1) = \frac{3}{4} - \frac{2k+3}{4*3^k} + \frac{k+1}{3^{k+1}} = \frac{3}{4} - \frac{2(k+1)+3}{4*3^{k+1}}$

This is as far as I could get! I need assistance simplifying the left hand side of this last line, assuming I have worked this through correctly. I have not worked too much with sequences so my trouble is mainly with getting a common denominator. Thanks for anyone that helps with this, I know it is a lot to read through.

2. The expression can be simplified by finding the lowest common denominator of $4\cdot 3^k$ and $3^{k+1}$.

3. That's what I was having problems with, is the lowest common denominator then $4*3^{k+1}$? And I need to take $3^{k+1} = (3^k)(3^1)$?

4. The lowest comon denominator is OK!

If you make the calculations carefully, you will have to compare

-((2k+1)+3) with -3(2k+3)+4(k+1)