Question

Find the sum of the series
$\frac{3\cdot 5}{5\cdot 10}+\frac{3\cdot 5\cdot 7}{5\cdot 10\cdot 15}+\frac{3\cdot 5\cdot 7\cdot 9}{5\cdot 10\cdot 15\cdot 20}+.....\infty$.

Answer 1

$\frac{3\cdot 5}{5\cdot 10}+\frac{3\cdot 5\cdot 7}{5\cdot 10\cdot 15}+\frac{3\cdot 5\cdot 7\cdot 9}{5\cdot 10\cdot 15\cdot 20}+.....\infty$

$n^{th}$ term of the above expansion.

$T_n=\frac{3\cdot 5\cdot 7\cdot .....\cdot (2n+3)}{5^{n+1}\cdot [(n+1)!]}$

$T_n=\frac{{\Pi }_{k=1}^n(2k+3)}{5^{n+1}(n+1)!}$

$T_n=\frac{2^n{\Pi }_{k=1}^n(k+\frac{3}{2})}{5^{n+1}(n+1)!}$

$T_n=\frac{2^n{\Pi }_{k=0}^{n-1}((k+1)+\frac{3}{2})}{5^{n+1}(n+1)!}$

$T_n=\frac{2^n{\Pi }_{k=0}^{n-1}(k+\frac{5}{2})}{5^{n+1}(n+1)!}$

$T_n={\left(\frac{2}{5}\right)}^n\frac{{\Pi }_{k=0}^{n-1}(k+\frac{5}{2})}{5(n+1)!}$

The expansion of ${(1+t)}^r$ is-

${(1+t)}^r={\sum }_0^{\infty }{t}^n\frac{{\Pi }_{k=0}^{n-1}(k+r)}{5(n+1)!}$

Here,

$t=\frac{2}{5}$

$r=\frac{5}{2}$

Therefore,

$\frac{3\cdot 5}{5\cdot 10}+\frac{3\cdot 5\cdot 7}{5\cdot 10\cdot 15}+\frac{3\cdot 5\cdot 7\cdot 9}{5\cdot 10\cdot 15\cdot 20}+.....\infty$

$={\sum }_{n=1}^{\infty }{T}_n$

$=1-{(1+\frac{2}{5})}^{\frac{2}{5}}$

$=1-{\left(\frac{7}{5}\right)}^{\frac{2}{5}}$