Question

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

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true. . .

Answer 1

Answer: (D)

We have $S=(a+b)+\left(a^2+ab+b^2\right)+\left(a^3+a^{2}b+a{b}^2+b^3\right)+\dots \dots \dots \dots \infty$
$=\frac{1}{a-b}\left[\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+\dots \dots \infty \right)$
$=\frac{1}{a-b}\left(\left(a^2+a^3+a^4+\dots \dots \dots \infty \right)-\left(b^2+b^3+\dots \dots \dots \right)\right)=\frac{1}{a-b}\left(\frac{a^2}{1-a}-\frac{b^2}{1-b}\right)$
$=\frac{1}{a-b}\left(\frac{a^2(1-b)-b^2(1-a)}{(1-a)(1-b)}\right)=\frac{a^2-b^2+a{b}^2-a^{2}b}{(a-b)(1-a)(1-b)}=\frac{(a-b)(a+b)+ab(b-a)}{(a-b)(1-a)(1-b)}=\frac{a+b-ab}{(1-a)(1-b)}$
$\therefore S_1$ is false and $S_2$ is true.