1

What is the step of solving this problem?

Evaluate: $$\sum_{i=1}^nia^i$$

For what I know, it's $$\sum_{i=1}^ni\times\sum_{i=1}^na^i$$

I know how to evaluate $$\sum_{i=1}^na^i=\frac{a(1-a^n)}{1-a}$$ but I am stuck on the the next steps.

Please help, thank you.

hlx98007
  • 149

2 Answers2

1

Just write the sum out. It's an arithmetico-geometric series. The combination of an arithmetic and geometric series.

Let $S_n =1a^1 + 2a^2 + 3a^3 + ... + na^n$ ... (1)

$aS_n = a^2 + 2a^3 + 3a^4 + ... + (n-1)a^n + na^{n+1}$ ... (2)

(1) - (2)

$S_n(1-a) = a^1 + a^2 + a^3 + ... + a^n - na^{n+1}$

Since $a^1 + a^2 + a^3 + ... + a^n = \frac{a(1-a^n)}{1-a}$ => geometric series

$S_n(1-a) = \frac{a(1-a^n)}{1-a} -na^{n+1}$

$S_n = \frac{a(1-a^n)}{(1-a)^2} - \frac{na^{n+1}}{1-a}$

This method I've used is useful for finding many formulas. Sometimes you just have to apply it a few times or split the summation into parts and apply it but it's pretty universal with series'

Here's the other proof you asked for.

Let $S_n = 1(2)^{n-1} + 4(2)^{n-2} + 9(2)^{n-3} + 16(2)^{n-4} + ... + (n^2-4n+4)(2)^3 +(n^2-2n+1)(2)^2 + n^2(2)^1$...(1)

I just took $(n-1)^2 = n^2 - 2n + 1$ and $(n-2)^2 = n^2-4n+4$

$2S_n = 2^n + 4(2)^{n-1} + 9(2)^{n-2} + 16(2)^{n-3} + ... + (n^2-2n+1)(2)^3 + n^2(2)^2$...(2)

(2) - (1)

$S_n = 1(2)^n + 3(2)^{n-1} + 5(2)^{n-2} + 7(2)^{n-3} + ... + (2n-3)(2)^3 + (2n-1)(2)^2 - n^2(2)^1$

Now since it has that inconvenient term at the end I'll separate the first part then put it back in later.

Let $M_n = 1(2)^n + 3(2)^{n-1} + 5(2)^{n-2} + 7(2)^{n-3} + ... + (2n-3)(2)^3 + (2n-1)(2)^2$..(3)

$2M_n = (2)^{n+1} + 3(2)^n + 5(2)^{n-1} + 7(2)^{n-2} + ... + (2n-3)(2)^4 + (2n-1)(2)^3$...(4)

(4) - (3)

$M_n = 2^{n+1} + 2(2)^n + 2(2)^{n-1} + 2(2)^{n-2} + ... + 2(2)^3 - (2n-1)(2)^2$

$M_n = 2^{n+1} + 2[(2)^n + (2)^{n-1} + (2)^{n-2} + ... + (2)^3] - (2n-1)(2)^2$

$M_n = 2^{n+1} + 2[\frac{2^n(1-(\frac{1}{2})^{n-3+1})}{1-\frac{1}{2}}] - (2n-1)(2)^2$

$M_n = 2^{n+1} + 2[\frac{2^n(1-(\frac{1}{2})^{n-2})}{\frac{1}{2}}] - (2n-1)(2)^2$

$M_n = 2^{n+1} + 4[{2^n(1-(\frac{1}{2})^{n-2})}] - (2n-1)(2)^2$

$S_n = M_n - 2(n)^2$

$S_n = 2^{n+1} + 4[{2^n(1-(\frac{1}{2})^{n-2})}] - (2n-1)(2)^2 - 2(n)^2$

I think that's right

StephanCasey
  • 1,240
  • 4
  • 14
  • 29
  • This is an easy and understandable approach to my problem and thank you for that! I have another series but this method seems not applicable. $$\sum_{i=1}^ni^2 2^{n-i}$$ – hlx98007 Sep 07 '14 at 12:59
  • Do you have an easier approach to this specific problem? I did my best with this method. It's not uncomplicated but would you expect it not to be with that summation. I edited the question – StephanCasey Sep 07 '14 at 14:19
1

$$\sum_{i=0}^n a^i=\frac{1-a^{n+1}}{1-a}$$ So $$\sum_{i=1}^n ia^i=a\cdot \frac{d}{da}\sum_{i=0}^n a^i=a\cdot \frac{d}{da}\left(\frac{1-a^{n+1}}{1-a}\right)=\cdots$$

idm
  • 11,824