# Series: Method of Difference

• May 28th 2011, 02:47 AM
Blizzardy
Series: Method of Difference
Given S = (4/5) + (7/5^2) + (10/5^3) + ... + (31/5^10), express S in Σ form.
By considering the difference between S and (1/5)S, find S.

So this is how I did it:
S = Σ (3r+1)/5^r (Upperlimit 10, Lower limit r = 1 )
(1/5)S = Σ (3r+1)/5^(r+1)

S - (1/5)S = Σ [(3r+1)/5^r - (3r+1)/5^(r+1)]
= + 5/4 - 4/25 r = 1
+ 1/25 - 7/125 r = 2
+ 10/125 - 10/625 r = 3
+ ...
I use the method of difference to solve but there don't seem to be any terms I can cancel away... Any idea what went wrong? (Thinking)
• May 28th 2011, 03:19 AM
FernandoRevilla
In general $\sum_{n=1}^{+\infty}(an+b)r^n=a\sum_{n=1}^{+\infty }nr^n+b\sum_{n=1}^{+\infty}r^n$. The second addend corresponds to a geometric series. For the first one and denoting $H=\sum_{n=1}^{+\infty}nr^n$

$\begin{Bmatrix} H=r+2r^2+3r^3+\ldots\\rH=r^2+2r^3+3r^4\ldots\end{m atrix}\Rightarrow (1-r)H=r+r^2+r^3+\ldots$

Now you can easily find $H$
• May 28th 2011, 04:42 AM
sa-ri-ga-ma
No nee to cancel.

S - (1/5)S = (4/5) + (7 - 4)/5^2 + (10 - 7)/5^3 + ............(31 - 28)/5^10 - 31/5^11

Can you proceed now?