We rewrite the series as the following:
∑ k = 0 ∞ ( 2 ( 3 4 ) k + 5 7 k ) = 2 ∑ k = 0 ∞ ( 3 4 ) k + 5 ∑ k = 0 ∞ ( 1 7 ) k {\displaystyle {\begin{aligned}\sum \limits _{k=0}^{\infty }\left(2\left({\frac {3}{4}}\right)^{k}+{\frac {5}{7^{k}}}\right)=2\sum \limits _{k=0}^{\infty }\left({\frac {3}{4}}\right)^{k}+5\sum \limits _{k=0}^{\infty }\left({\frac {1}{7}}\right)^{k}\end{aligned}}}
each of the summations is a Geometric Series, and we can use the following formula for the sum of a geometric series:
∑ k = 0 ∞ a r k = a 1 − r {\displaystyle \sum \limits _{k=0}^{\infty }ar^{k}={\frac {a}{1-r}}}
So, we have:
∑ k = 0 ∞ ( 3 4 ) k = 1 + 3 4 + ( 3 4 ) 2 + . . . = 1 1 − 3 4 = 4 , {\displaystyle \sum \limits _{k=0}^{\infty }\left({\frac {3}{4}}\right)^{k}=1+{\frac {3}{4}}+({\frac {3}{4}})^{2}+...={\frac {1}{1-{\frac {3}{4}}}}=4,}
and
∑ k = 0 ∞ ( 1 7 ) k = 1 + 1 7 + ( 1 7 ) 2 + . . . = 1 1 − 1 7 = 7 6 . {\displaystyle \sum \limits _{k=0}^{\infty }\left({\frac {1}{7}}\right)^{k}=1+{\frac {1}{7}}+({\frac {1}{7}})^{2}+...={\frac {1}{1-{\frac {1}{7}}}}={\frac {7}{6}}.}
And, we finally get
∑ k = 0 ∞ ( 2 ( 3 4 ) k + 5 7 k ) = 2 ∑ k = 0 ∞ ( 3 4 ) k + 5 ∑ k = 0 ∞ ( 1 7 ) k = ( 2 × 4 ) + ( 5 × 7 6 ) = 83 6 . {\displaystyle \sum \limits _{k=0}^{\infty }\left(2\left({\frac {3}{4}}\right)^{k}+{\frac {5}{7^{k}}}\right)=2\sum \limits _{k=0}^{\infty }\left({\frac {3}{4}}\right)^{k}+5\sum \limits _{k=0}^{\infty }\left({\frac {1}{7}}\right)^{k}=(2\times 4)+(5\times {\frac {7}{6}})={\frac {83}{6}}.}
