Question

$\mathrm{Find \: the \: sum \: of \: the \: infinite \: \: series \:} \\ \\ { \frac{3}{4} + \frac{3.5}{4.8} + \frac{3.5.7}{4.8.12} +...... \infty } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 1

$\mathrm \red{Let \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ { \: \: S = \frac{3}{4} + \frac{3.5}{4.8} + \frac{3.5.7}{4.8.12} +...... }$

$\mathrm{Add \: '1' \: on \: both \: sides}$

$\\ ⇒1+S \: = 1 + \frac{3}{4} + \frac{3.5}{4.8} + \frac{3.5.7}{4.8.12} +..... \: \: \: \: \: \: \: \: \: \: \\ \\ { = 1 + \frac{3}{1} \bigg( \frac{1}{4} \bigg) + \frac{3.5}{1.2} { \bigg( \frac{1}{4} \bigg)}^{2} + \frac{3.5.7}{1.2.3} { \bigg( \frac{1}{4} \bigg)}^{3} + ....} \\ \\ \mathrm{ = 1 + \frac{3}{1!} \bigg( \frac{1}{4} \bigg)+ \frac{3.5}{2!} { \bigg( \frac{1}{4} \bigg)}^{2} + \frac{3.5.7}{3!} { \bigg( \frac{1}{4} \bigg)}^{3} + ...} \\ \\ \mathrm{R.H.S \: \: is \: compared \: with = 1 + \frac{p}{1!} \bigg( \frac{x}{q} \bigg)+ \frac{p(p + q)}{2!} { \bigg( \frac{x}{q} \bigg)}^{2} + ....}$

$\\   \mathrm{Where  \: \:  p = 3;  \: p+q = 5  \: ⇒ q = 2  \:  \: and } \\ \\   \mathrm{ \frac{x}{q}  =  \frac{1}{4}⇒x =  \frac{2}{4}  =  \frac{1}{2}  } \\  \\    \mathrm{\because  1 +  \frac{p}{1!} \bigg( \frac{x}{q} \bigg)+  \frac{p(p + q)}{2!}  { \bigg(  \frac{x}{q}  \bigg)}^{2} + .... = {(1 - x)}^{ \frac{ - p}{q} } } \\ \\ \mathrm{ \therefore \: 1+S = {\bigg( 1 - \frac{1}{2} \bigg)}^{ \frac{ - 3}{2} }} \\ \\ \mathrm{1+S \: = \: { \bigg( \frac{1}{2} \bigg) }^{ \frac{ - 3}{2} } } \\ \\ \mathrm{1+S \: = \: {(2)}^{3/2} } \\ \\ \mathrm{S \: = \: \sqrt{8} - 1} \\ \\ \boxed{ \red{\frac{3}{4} + \frac{3.5}{4.8} + \frac{3.5.7}{4.8.12} +...... \infty \: = \: \sqrt{8} - 1}}$

Answer 2

$\mathrm{Find \: the \: sum \: of \: the \: infinite \: \: series \:} \\ \\ { \frac{3}{4} + \frac{3.5}{4.8} + \frac{3.5.7}{4.8.12} +...... \infty } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$