The sum to infinity of the following series 2 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) Given series
= $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + .......\infty $

Vedclass · Accepted Answer

$7/2$

Vedclass · Answer

(c) Given series
= $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + .......\infty $

$ = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ......\infty } \right)$
+ $\left( {1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ......\infty } \right)$

$ = \left( {\frac{1}{{1 - (1/2)}}} \right) + \left( {\frac{1}{{1 - (1/3)}}} \right) = 2 + \frac{3}{2} = \frac{7}{2}$.

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + ........$, will be

Similar Questions

$0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = $

The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + .........$ is

Find four numbers forming a geometric progression in which the third term is greater than the first term by $9,$ and the second term is greater than the $4^{\text {th }}$ by $18 .$

If $x, {G_1},{G_2},\;y$ be the consecutive terms of a $G.P.$, then the value of ${G_1}\,{G_2}$ will be

The sum of the series $5.05 + 1.212 + 0.29088 + ...\,\infty $ is