1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement-1: If | a |<1 and | b |<1 then (a+b)+(a2+a b+b2)+(a3+a2 b+a b2+b3)+ ldots ldots ldots ldots=(a+b+a b/(1-a)(1-b)) Statement-2: a+a r+a r2+ ldots ldots . . .=(a/1-r),-1< r< 1 .

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Q. Statement-1: If $| a |<1$ and $| b |<1$ then $(a+b)+\left(a^2+a b+b^2\right)+\left(a^3+a^2 b+a b^2+b^3\right)+\ldots \ldots \ldots \ldots=\frac{a+b+a b}{(1-a)(1-b)} $
Statement-2: $a+a r+a r^2+\ldots \ldots . . .=\frac{a}{1-r},-1< r< 1 .$

Sequences and Series

Solution:

We have $S =( a + b )+\left( a ^2+ ab + b ^2\right)+\left( a ^3+ a ^2 b + ab ^2+ b ^3\right)+\ldots \ldots \ldots \ldots \infty$
$=\frac{1}{a-b}\left[\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+\ldots \ldots \infty\right)$
$=\frac{1}{a-b}\left(\left(a^2+a^3+a^4+\ldots \ldots \ldots \infty\right)-\left(b^2+b^3+\ldots \ldots \ldots\right)\right)=\frac{1}{a-b}\left(\frac{a^2}{1-a}-\frac{b^2}{1-b}\right)$
$=\frac{1}{a-b}\left(\frac{a^2(1-b)-b^2(1-a)}{(1-a)(1-b)}\right)=\frac{a^2-b^2+a b^2-a^2 b}{(a-b)(1-a)(1-b)}=\frac{(a-b)(a+b)+a b(b-a)}{(a-b)(1-a)(1-b)}=\frac{a+b-a b}{(1-a)(1-b)}$
$\therefore S _1$ is false and $S _2$ is true.