Question

If $|a|<1$ and $|b|<1$, then the sum of the series
$1+(1+a)b+(1+a+a^2)b^2+(1+a+a^2+a^3)b^3+\cdots$ is

• A. $\frac{1}{(1-a)(1-b)}$
• B. $\frac{1}{(1-b)(1-ab)}$
• C. $\frac{1}{(1-a)(1-b)}$
• D. $\frac{1}{(1-a)(1-b)(1-ab)}$

Answer 1

The correct option is B $\frac{1}{(1-b)(1-ab)}$

$1+(1+a)b+(1+a+a^2)b^2+(1+a+a^2+a^3)b^3+\cdots =\sum _{n=1}^{\infty }(1+a+a^2+\cdots +a^{n-1})b^{n-1}=\sum _{n=1}^{\infty }\left(\frac{1-a^n}{1-a}\right)b^{n-1}=\sum _{n=1}^{\infty }\frac{b^{n-1}}{1-a}-\sum _{n=1}^{\infty }\frac{a^n{b}^{n-1}}{1-a}=\frac{1}{1-a}\sum _{n=1}^{\infty }{b}^{n-1}-\frac{a}{1-a}\sum _{n=1}^{\infty }(ab{)}^{n-1}=\frac{1}{1-a}\times \frac{1}{1-b}-\frac{a}{1-a}\times \frac{1}{1-ab}=\frac{1}{(1-b)(1-ab)}$