# Math Help - find sum

1. ## find sum

$(\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{ 3^9}+\frac{4}{3^{10}})+.....$

2. Originally Posted by perash
$(\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{ 3^9}+\frac{4}{3^{10}})+.....$
Put each of the bracketed terms over a common denominator and simplify a bit. That will leave you with a series you should recognise and be able to sum.

RonL

3. Originally Posted by perash
$(\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{ 3^9}+\frac{4}{3^{10}})+.....$
Off topic, I know, but this is a very cool series

$T_{n} = \frac{n}{3^{2n + 1}} + \frac{n}{3^{2n + 2}}$

4. Originally Posted by CaptainBlack
Put each of the bracketed terms over a common denominator and simplify a bit. That will leave you with a series you should recognise and be able to sum.

RonL
i tried that and ended up with $\frac 4{3^2} \sum_{n = 1}^{\infty} \frac n{9^n}$. but i can't remember for the life of me how to sum something like that...

5. Originally Posted by Jhevon
i tried that and ended up with $\frac 4{3^2} \sum_{n = 1}^{\infty} \frac n{9^n}$. but i can't remember for the life of me how to sum something like that...
Put:

$S(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$

Now differentiate term by term:

$S'(x)=\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^2}$

so:

$xS'(x)=\sum_{n=1}^{\infty} n x^n=\frac{x}{(1-x)^2}$

Now put $x=1/9$.

(also see attachment)

RonL

6. Originally Posted by CaptainBlack
Put:
$S(x)=\sum_{n=1}^{\infty} x^n=\frac{1}{1-x}$
Hmm.. Wasn't that $\frac{x}{1-x}?$

7. Without differentiating

We want to find $G(x)=\sum_{k=0}^{\infty}{\left(k\cdot{x^k}\right)}$ ( By the root test you can see that the series converges for $|x|<1$)

Now we consider:
$x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^{k +1}}\right)}=\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}$

So: $G(x)-x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^k} \right)}-\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}=\sum_{k=1}^{\infty}{(k-k+1)\cdot{x^k}}=\sum_{k=1}^{\infty}{x^k}=\frac{x}{ 1-x}$ whenever $|x|<1$ (Because it is a geometric series)

Thus: $(1-x)\cdot{G(x)}=\frac{x}{1-x}$ and we obtain: $G(x)=\frac{x}{(1-x)^2}$

8. Originally Posted by wingless
Hmm.. Wasn't that $\frac{x}{1-x}?$
You are right, but luckily we get the same result because the constant 1 vanishes when differentiating

$(1+x+x^2+...)'=(x+x^2+...)'$

9. Hello, perash!

$S \;=\;\left(\frac{1}{3^3}+\frac{1}{3^4}\right)+\lef t(\frac{2}{3^5}+\frac{2}{3^6}\right)+\left(\frac{3 }{3^7}+\frac{3}{3^8}\right)+\left(\frac{4}{3^9}+\f rac{4}{3^{10}}\right)+ \cdots$

We have: . . . . . . . . . . $S \;=\;\frac{4}{3^4} + \frac{8}{3^6} + \frac{12}{3^8} + \frac{16}{3^{10}} + \cdots$

Factor: . . . . . . . . . . . $S\;=\;\frac{4}{3^4}\left[1 + \frac{2}{3^2} + \frac{3}{3^4} + \frac{4}{3^4} + \cdots\right]\;\;{\color{blue}[1]}$

Multiply by $\frac{1}{3^2}:$ . . . . . $\frac{1}{9}S \;=\;\frac{4}{3^4}\left[\qquad \frac{1}{3^2} + \frac{2}{3^4} + \frac{3}{3^6} + \cdots\right]\;\;{\color{blue}[2]}$

Subtract [2] from [1]: . $\frac{8}{9}S \;=\;\frac{4}{3^4}\underbrace{\left[1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \cdots\right]}_{\text{geometric series: }a = 1,\:r = 1/9}$

Hence, we have: . . . . $\frac{8}{9}S \;=\;\frac{4}{81}\left[\frac{9}{8}\right] \quad\Rightarrow\quad\boxed{S \:=\:\frac{1}{16}}$

10. Originally Posted by wingless
Hmm.. Wasn't that $\frac{x}{1-x}?$
No the lower limit of summation should have been 0, and now is (it is just a constant term that
disappears on differentiation, or sometimes when just typing).

RonL

11. This wonderful solution

thank you Soroban