# Series

• Dec 22nd 2009, 01:59 AM
arze
Series
Find the sum S(x) of the series $1+2x+3x^2+...+(n+1)x^n$ by finding $(1-x)S(x)$.

I multiplies the series by (1-x) and got $1+x+x^2+...+x^{n-1}$
And the sum of this would be $\frac{1(1-x^n)}{(1-x)}$
Then what would S(x) be? $\frac{1(1-x^n)}{(1-x)}.\frac{1}{(1-x)}$?
The answer is supposed to be $\frac{[1-(n+2)x^{n+1}+(n+1)x^{n+2}]}{(1-x)^2}$
Thanks
• Dec 22nd 2009, 04:40 AM
mr fantastic
Quote:

Originally Posted by arze
Find the sum S(x) of the series $1+2x+3x^2+...+(n+1)x^n$ by finding $(1-x)S(x)$.

I multiplies the series by (1-x) and got $1+x+x^2+...+x^{n-1}$ Mr F says: This is wrong. It's ${\color{red}1+x+x^2+...+x^n - (n+1) x^{n+1}}$.

And the sum of this would be $\frac{1(1-x^n)}{(1-x)}$
Then what would S(x) be? $\frac{1(1-x^n)}{(1-x)}.\frac{1}{(1-x)}$?
The answer is supposed to be $\frac{[1-(n+2)x^{n+1}+(n+1)x^{n+2}]}{(1-x)^2}$
Thanks

..
• Dec 22nd 2009, 05:52 PM
arze
So i find the sum of the series $x^n-(n+1)x^{n+1}$?
• Dec 23rd 2009, 02:11 AM
mr fantastic
Quote:

Originally Posted by arze
So i find the sum of the series $x^n-(n+1)x^{n+1}$?

No.

$(1 - x) S(x) = {\color{red}1+x+x^2+...+x^n} - (n+1) x^{n+1}$.

The red stuff is a geometric series - use the usual rule to get an expression for it. Then make S(x) the subject and simplify the result (simplifying should be easy since you know the answer to aim for).
• Dec 23rd 2009, 08:27 AM
Soroban
Hello, arze!

Quote:

Find the sum: . $S(x) \:=\:1+2x+3x^2+...+(n+1)x^n$ .by finding $(1-x)S(x)$

The answer is: . $S(x) \:=\:\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(1-x)^2}$

$\text{We have: }\qquad\qquad\;\; S(x) \;=\; 1 + 2x + 3x^2 + 4x^2 + \hdots + nx^{n-1} + (n+1)x^n$

$\text{Multiply by }x\!:\qquad xS(x) \;=\; \qquad x + 2x^2 + 3x^3 + 4x^4 + \quad\hdots$ . . $+ nx^n + (n+1)x^{n+1}$

$\text{Subtract: }\quad S(x) - xS(x) \;=\;1 + x + x^2 + x^3 + \hdots + x^n - (n+1)x^{n+1}$

$\text{We have: }\;(1-x)\cdot S(x) \;=\;\underbrace{1 + x^2+x^3+ \hdots + x^n}_{\text{geometric series}} - \,(n+1)x^{n+1}$ .[1]

. . The geometric series has sum: . $\frac{1-x^{n+1}}{1-x}$

So [1] becomes: . $(1-x)\cdot S(x) \;=\;\frac{1-x^{n+1}}{1-x} - (n+1)x^{n+1} \;=$ . $\frac{1 - x^{n+1} - (n+1)x^{n+1} + (n+1)x^{n+2}}{1-x}$

Hence: . $(1-x)\cdot S(x) \;=\;\frac{1-(n+2)x^{n+1} + (n+1)x^{n+2}}{1-x}$

Therefore: . $S(x) \;=\;\frac{1 - (n+2)x^{n+1} + (n+1)x^{n+2}}{(1-x)^2}$