(13+47)+(13^2+1347+4^27^2)+(13^3+13^247+134^27^2+4^37^3)+ up to infinite terms, is equal t…

Question

$\left(\dfrac13+\dfrac47\right)+\left(\dfrac{1}{3^2}+\dfrac13\cdot\dfrac47+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\cdot\dfrac47+\dfrac13\cdot\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+\cdots$ up to infinite terms, is equal to:

Accepted Answer

Let $a=\dfrac13,\ b=\dfrac47$. The $n$-th bracket is
$$T_n=a^n+a^{n-1}b+a^{n-2}b^2+\cdots+b^n=a^n\left[\frac{(b/a)^{n+1}-1}{(b/a)-1}\right],$$
the sum of $n+1$ terms of a G.P. with ratio $b/a=\dfrac{12}{7}$.

Simplifying (using $b/a-1=5/7$):
$$T_n=\frac{7}{5}\left[a^n\left(\frac ba\right)^{n+1}\cdot\frac ab-a^n\right]\cdot\ (\text{after algebra})=\frac{12}{5}\left(\frac47\right)^n-\frac75\left(\frac13\right)^n.$$

$$S_\infty=\sum_{n=1}^\infty T_n=\frac{12}{5}\sum_{n=1}^\infty\left(\frac47\right)^n-\frac75\sum_{n=1}^\infty\left(\frac13\right)^n=\frac{12}{5}\cdot\frac{4/7}{1-4/7}-\frac75\cdot\frac{1/3}{1-1/3}$$

$$=\frac{12}{5}\cdot\frac43-\frac75\cdot\frac12=\frac{16}{5}-\frac{7}{10}=\frac{32-7}{10}=\frac{25}{10}=\frac52.$$

Solutions · 1