Sum the series

Sn = a + (a+d)r + (a+2d)r^2 + ... + [a+(n-1)d]r^(n-1)

n-1

Sum (a+nd)r^n

n=0

Presumably I need to split this up into a sum of ar^n and ndr^n

I have already summed ar^n, how would I go about summing ndr^n ?

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- Oct 23rd 2005, 08:41 AMSunGodSumming a series.
Sum the series

Sn = a + (a+d)r + (a+2d)r^2 + ... + [a+(n-1)d]r^(n-1)

n-1

Sum (a+nd)r^n

n=0

Presumably I need to split this up into a sum of ar^n and ndr^n

I have already summed ar^n, how would I go about summing ndr^n ? - Oct 24th 2005, 04:31 PMhpeQuote:

Originally Posted by**SunGod**

$\displaystyle S_0 = a + ar + ar^2 + \dots + ar^n = a\frac{r^{n+1}-1}{r-1}$

$\displaystyle S_1 = dr + dr^2 + \dots + dr^n = dr\frac{r^n-1}{r-1} = d \frac{r^{n+1}-r}{r-1}$

$\displaystyle S_2 = dr^2 + \dots + dr^n = dr^2\frac{r^{n-1}-1}{r-1} = d\frac{r^{n+1}-r^2}{r-1}$

and so on until

$\displaystyle S_n = dr^n = d \frac{r^{n+1}-r^n}{r-1}$

Then add all these sums, using the formula for geometric sums. There is also a way to do this with calculus. - Oct 25th 2005, 07:24 AMSunGod
Hey thanks for the solution, I've already handed in the work alas without the proof of that question, but at least now I know the logic behind working out a complicated summation like that (complicated on my level, anyway)

I've also read the other solution which requires calculus, thanks alot :)