((1)/(3)+(4)/(7))+(13^2+(1)/(3) (4)/(7)+4^27^2)+(13^3+13^2 (4)/(7)+(1)/(3) 4^27^2+4^37^3)+…

Question

$(\frac{1}{3}+\frac{4}{7})+(\frac{1}{{3}^{2}}+\frac{1}{3}	imes \frac{4}{7}+\frac{{4}^{2}}{{7}^{2}})+(\frac{1}{{3}^{3}}+\frac{1}{{3}^{2}}	imes \frac{4}{7}+\frac{1}{3}	imes \frac{{4}^{2}}{{7}^{2}}+\frac{{4}^{3}}{{7}^{3}})+\ldots$ upto infinite terms, is equal to

Accepted Answer

Let $a=\frac{1}{3}$ and $b=\frac{4}{7}$

So Question is

${S}_{\infty}=(a+b)+({a}^{2}+ab+{b}^{2})+({a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3})+\ldots \infty$

So clearly each bracket is in G.P. having ( $n+1$ ) terms.

So general term is

${T}_{n}=({a}^{n}+{a}^{n-1}b+{a}^{n-2}{b}^{2}\cdots +{b}^{n})$

${T}_{n}$ is G.P. first term ${a}^{n}$, common ratio $=\frac{b}{a},(n+1)$ terms

${T}_{n}={a}^{n}[\frac{{(\frac{b}{a})}^{n+1}-1}{\frac{b}{a}-1}]$

Now substitute $a=\frac{1}{3}$ and $b=\frac{4}{7}$

$\begin{gathered} {T}_{n}={(\frac{1}{3})}^{n}[\frac{{(\frac{12}{7})}^{n+1}-1}{\frac{12}{7}-1}] \ {T}_{n}={(\frac{1}{3})}^{n}\frac{[{(\frac{12}{7})}^{n}\cdot \frac{12}{7}-1]\cdot 7}{5} \ {T}_{n}={(\frac{1}{3})}^{n}{(\frac{12}{7})}^{n}\cdot \frac{12}{7}	imes \frac{7}{5}-1	imes \frac{7}{5}	imes {(\frac{1}{3})}^{n} \ {T}_{n}=\frac{12}{5}{(\frac{4}{7})}^{n}-\frac{7}{5}{(\frac{1}{3})}^{n} \end{gathered}$

Now Required sum $({S}_{\infty})=\sum _{\infty}^{n=1}{T}_{n}$

$\begin{gathered} {S}_{\infty} =\sum _{n=1}^{\infty}\frac{12}{5}{(\frac{4}{7})}^{n}-\frac{7}{5}{(\frac{1}{3})}^{n} \ {S}_{\infty} =\frac{12}{5}\sum _{n=1}^{\infty}{(\frac{4}{7})}^{n}-\frac{7}{5}\sum _{n=1}^{\infty}{(\frac{1}{3})}^{n} \ {S}_{\infty} =\frac{12}{5}[\frac{4}{7}+{(\frac{4}{7})}^{2}+{(\frac{4}{7})}^{3}\ldots ]-\frac{7}{5}[\frac{1}{3}+{(\frac{1}{3})}^{2}+{(\frac{1}{3})}^{3}\ldots ] \end{gathered}$

Both series are infinite G.P.

$\begin{gathered} {S}_{\infty}=\frac{12}{5}[\frac{4/7}{1-4/7}]-\frac{7}{5}[\frac{1/3}{1-1/3}] \ =\frac{12}{5}[\frac{4}{3}]-\frac{7}{5}[\frac{1}{2}] \ =\frac{16}{5}-\frac{7}{10}=\frac{32-7}{10}=\frac{25}{10}=\frac{5}{2} \end{gathered}$

Solutions · 1