Prove that:

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

$\displaystyle P(1) = \frac{1}{3^1} = \frac{3}{4} - \frac{5}{4*3^1} $

$\displaystyle \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,

$\displaystyle 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

$\displaystyle 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:

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

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.