Discussion

Explanation:

S = $\frac{3}{7}$ + $\frac{4}{7^2}$ + $\frac{5}{7^3}$ + $\frac{3}{7^4}$ + $\frac{4}{7^5}$ + $\frac{5}{7^6}$ + ...

Now here S is a summation of 3 distinct infinite geometric series
And we know that sum of an infinite geometric progression = (a/1-r) where a is the first term and r is the common ratio .

Therefore we can say S will be:

$\frac{3}{{\displaystyle \frac{7}{1-{\displaystyle \frac{1}{7^3}}}}}$ + $\frac{4}{{\displaystyle \frac{7^2}{1-{\displaystyle \frac{1}{7^3}}}}}$ + $\frac{5}{{\displaystyle \frac{7^3}{1-{\displaystyle \frac{1}{7^3}}}}}$

solving we get S = $\frac{180}{7^3-1}$ = $\frac{180}{342}$

» Your doubt will be displayed only after approval.