Sum to n groups the following series
$\displaystyle (x+y) + (x^2+xy+y^2) + {x^3+(x^2)*y+x(y^2)+y^3}$
Need help fast..............Thanks a lot in advance.
No I did not mean that.........
I tried till
$\displaystyle (x+y) + (x(x+y) + y^2) + (x(x^2+xy+y^2) + y^3) + .......$
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.......
NO.....It's not......but see....
$\displaystyle (x+y) + (x(x+y) + y^2) + (x^2(x+y) + y^2(x+y)) +......$
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.
$\displaystyle x^{4}-y^{4} = (x-y)(x+y)(x^{2}+y^{2})$
therefore, Series reduces to,
$\displaystyle (x^{2}-y^{2})/(x-y) + (x^{3}-y^{3})/(x-y) + (x^{4}-y^{4})/(x-y)+....$
Simplify further, you'll obtain something resembling a geometric series
Obviously this is under the assumption that x and y are not equal.
Hello, Arka!
Sum to $\displaystyle \,n$ groups the following series:
. . $\displaystyle S_n \;=\;(x+y) + (x^2+xy+y^2) + (x^3 + x^2y+xy^2+y^3) + \hdots$
Multiply each group by $\displaystyle \frac{x-y}{x-y}$
. . $\displaystyle \begin{array}{ccccccc}
\dfrac{x+y}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^2-y^2}{x-y} \\ \\[-3mm]
\dfrac{x^2+xy+y^2}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^3-y^3}{x-y} \\ \\[-3mm]
\dfrac{x^3+x^2y+xy^2+y^3}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^4 - y^4}{x-y} \\ \\[-3mm]
\vdots && \vdots \\
\dfrac{x^n + x^{n-1}y + \hdots + xu^{n-1} + y^n}{1}\cdot\dfrac{x-y}{x-y} &=& \dfrac{x^{n+1}-y^{n+1}}{x-y} \end{array}$
$\displaystyle \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} $
$\displaystyle \text{The numerator has two geometric series:}$
. . $\displaystyle S_x \:=\:x^2\cdot\dfrac{1-x^n}{1-x}\;\text{ and }\;S_y \;=\;y^2\cdot\dfrac{1-y^n}{1-y}$
$\displaystyle \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} $