# Math Help - Series

1. ## 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

2. 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
..

3. So i find the sum of the series $x^n-(n+1)x^{n+1}$?

4. 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).

5. Hello, arze!

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