If |x|

Question

(A) Given the series $S = (x+y) + (x^2+xy+y^2) + (x^3+x^2y+xy^2+y^3) + \dots$
Multiply and divide by $(x-y)$:
$S = \frac{(x-y)(x+y) + (x-y)(x^2+xy+y

Vedclass · Accepted Answer

$\frac{x+y-xy}{(1-x)(1-y)}$

Vedclass · Answer

(A) Given the series $S = (x+y) + (x^2+xy+y^2) + (x^3+x^2y+xy^2+y^3) + \dots$
Multiply and divide by $(x-y)$:
$S = \frac{(x-y)(x+y) + (x-y)(x^2+xy+y^2) + (x-y)(x^3+x^2y+xy^2+y^3) + \dots}{x-y}$
Using the identity $(x-y)(x^n + x^{n-1}y + \dots + y^n) = x^{n+1} - y^{n+1}$:
$S = \frac{(x^2-y^2) + (x^3-y^3) + (x^4-y^4) + \dots}{x-y}$
$S = \frac{(x^2+x^3+x^4+\dots) - (y^2+y^3+y^4+\dots)}{x-y}$
Using the sum of infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$:
$S = \frac{\frac{x^2}{1-x} - \frac{y^2}{1-y}}{x-y} = \frac{x^2(1-y) - y^2(1-x)}{(1-x)(1-y)(x-y)}$
$S = \frac{x^2 - x^2y - y^2 + xy^2}{(1-x)(1-y)(x-y)} = \frac{(x^2-y^2) - xy(x-y)}{(1-x)(1-y)(x-y)}$
$S = \frac{(x-y)(x+y) - xy(x-y)}{(1-x)(1-y)(x-y)} = \frac{(x-y)(x+y-xy)}{(1-x)(1-y)(x-y)}$
$S = \frac{x+y-xy}{(1-x)(1-y)}$

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is:

Similar Questions

The first term of a $G.P.$ whose second term is $2$ and sum to infinity is $8$,will be

The sum of the series $(1^2 + 1) \cdot 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n!$ is:

Given $a_1, a_2, a_3, \dots$ form an increasing geometric progression with common ratio $r$ such that $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2014$,then the number of ordered pairs of integers $(a_1, r)$ is equal to

Let $f(n) = [\frac{1}{3} + \frac{3n}{100}]n$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then $\sum_{n=1}^{56} f(n)$ is equal to

If $m$ arithmetic means $(A.Ms)$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that the $4^{\text{th}}$ $A.M.$ is equal to the $2^{\text{nd}}$ $G.M.$,then $m$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module