Question

If $|a|<1,|b|<1$ then the sum of the series $1+(1+a)b+(1+a+a^2)b^2+(1+a+a^2+a^3)b^3+$ ………… to infinite terms is

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

Answer 1

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

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