I am supposed to find a formula for the sum r+2r^2 + 3r^3 + ... + nr^n. I know that it's the summation of k=1 to n of kr^k, but I don't know the equivalent formula to that. Can anyone help?
What have you tried? It looks trickier than it is.
You remember this one, right? the sum r + r^2 + r^3 + ... + r^n
Surprisingly, it's almost no more difficult.
Your series:
r + 2r^2 + 3r^3 + ... + nr^n
Give it a Sum
1) r + 2r^2 + 3r^3 + ... + nr^n = S
Multiply by r
2) r^2 + 2r^3 + 3r^4 + ... + nr^(n+1) = Sr
Subtract 2) from 1)
r + r^2 + 3r^3 + ... + nr^n - nr^(n+1) = S - Sr
It doesn't look like we're getting anywhere, does it? Before we give up, let's just group it a little.
[r + r^2 + 3r^3 + ... + nr^n] - nr^(n+1) = S - Sr
What does that part in the brackets look like? Familiar?
You have to reach out and grab these. Don't just answer the question. Think about what the answer might mean and if there are any important implications?
What is the formula for the sum in the brackets? Is it a sufficiently simple expression that you could substitute in the brackets and still manage to solve for S?
Go for it!
We have the identity:
$\displaystyle \displaystyle 1+x+x^2+x^3+\ldots +x^n=\frac{x^{n+1}-1}{x-1}$
Take derivative of both sides:
$\displaystyle \displaystyle 1+2x+3x^2+\ldots+nx^{n-1}=\left(\frac{x^{n+1}-1}{x-1}\right)'=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$
Now multiply both sides by $\displaystyle x$ and you're done.