# Thread: Sum of Geometric series........

1. ## Sum of Geometric series........

Sum to n groups the following series

$(x+y) + (x^2+xy+y^2) + {x^3+(x^2)*y+x(y^2)+y^3}$

Need help fast..............Thanks a lot in advance.

2. Sum to n groups the following series

$(x+y) + (x^2+xy+y^2) + (x^3+(x^2)*y+x(y^2)+y^3)$

Need help fast..............Thanks a lot in advance.[/QUOTE]

Sorry..... I forgot to give brackets in the 3rd group........

3. And what have you done or tried so far?

4. Originally Posted by Arka
Sum to n groups the following series

$(x+y) + (x^2+xy+y^2) + (x^3+(x^2)*y+x(y^2)+y^3)$

Need help fast..............Thanks a lot in advance.

sure you didn't mean ...

$(x+y) + (x^2 + 2xy + y^2) + (x^3 + 3x^2y + 3xy^2 + y^3) + ...$

???

5. No I did not mean that.........

I tried till

$(x+y) + (x(x+y) + y^2) + (x(x^2+xy+y^2) + y^3) + .......$

Now, from here if I get the sum of the nth group , I can easily find the sum to n groups.............but that's what I'm having trouble finding.......

6. Originally Posted by skeeter
sure you didn't mean ...
$(x+y) + (x^2 + 2xy + y^2) + (x^3 + 3x^2y + 3xy^2 + y^3) + ...$???
Originally Posted by Arka
No I did not mean that.........
Then that is not a geometric series.

7. NO.....It's not......but see....

$(x+y) + (x(x+y) + y^2) + (x^2(x+y) + y^2(x+y)) +......$

I definitely see a pattern up there... and I belive it is possible to find the sum of the nth group and then the sum to n groups.........but I can't figure out how to do it........

And the chapter in which this sum is given is called "Geometrical Progression".......so I believe that even if the series as a whole is not in G.P. some part of it must be in G.P.

8. $x^{4}-y^{4} = (x-y)(x+y)(x^{2}+y^{2})$

therefore, Series reduces to,

$(x^{2}-y^{2})/(x-y) + (x^{3}-y^{3})/(x-y) + (x^{4}-y^{4})/(x-y)+....$

Simplify further, you'll obtain something resembling a geometric series

Obviously this is under the assumption that x and y are not equal.

9. How do I simplify further.......?

10. See the common denominator,

u have, $(1/(x-y))((x^{2}+x^{3}+......)-(y^{2}+y^{3}+......))$ upto n terms

so there are two geometric progressions in x and y. Now sum each upto n terms

11. Hello, Arka!

Sum to $\,n$ groups the following series:

. . $S_n \;=\;(x+y) + (x^2+xy+y^2) + (x^3 + x^2y+xy^2+y^3) + \hdots$

Multiply each group by $\frac{x-y}{x-y}$

. . $\begin{array}{ccccccc}
\dfrac{x+y}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^2-y^2}{x-y} \\ \\[-3mm]
\dfrac{x^2+xy+y^2}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^3-y^3}{x-y} \\ \\[-3mm]
\dfrac{x^3+x^2y+xy^2+y^3}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^4 - y^4}{x-y} \\ \\[-3mm]
\vdots && \vdots \\
\dfrac{x^n + x^{n-1}y + \hdots + xu^{n-1} + y^n}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^{n+1}-y^{n+1}}{x-y} \end{array}$

$\text{Hence: }\;S_n \;=\;\dfrac{(x^2+x^3+x^4 + \hdots + x^{n+1}) - (y^2 + y^3 + y^4 + \hdots + y^{n+1})}{x-y}$

$\text{The numerator has two geometric series:}$

. . $S_x \:=\:x^2\cdot\dfrac{1-x^n}{1-x}\;\text{ and }\;S_y \;=\;y^2\cdot\dfrac{1-y^n}{1-y}$

$\text{Therefore: }\;S_n \;=\;\dfrac{x^2\cdot\frac{1-x^n}{1-x} - y^2\cdot\frac{1-y^n}{1-y}}{x-y}$