Problem: The sum to infinity of 17+272+173+274+…\dfrac{1}{7}+\dfrac{2}{7^{2}}+\dfrac{1}{7^{3}}+\dfrac{2}{7^{4}}+\ldots71+722+731+742+… is:
Answer Choices:
A. 15\dfrac{1}{5}51
B. 124\dfrac{1}{24}241
C. 548\dfrac{5}{48}485
D. 116\dfrac{1}{16}161
E. none of these answers Solution:
Rearrange the terms:
s1=17+173+⋯=1/71−(1/72)=748s2=272+274+⋯=2/721−(1/72)=248s=s1+s2=316\begin{aligned} & s_{1}=\dfrac{1}{7}+\dfrac{1}{7^{3}}+\cdots=\dfrac{1 / 7}{1-\left(1 / 7^{2}\right)}=\dfrac{7}{48} \\ & s_{2}=\dfrac{2}{7^{2}}+\dfrac{2}{7^{4}}+\cdots=\dfrac{2 / 7^{2}}{1-\left(1 / 7^{2}\right)}=\dfrac{2}{48} \\ & s=s_{1}+s_{2}=\dfrac{3}{16} \end{aligned} s1=71+731+⋯=1−(1/72)1/7=487s2=722+742+⋯=1−(1/72)2/72=482s=s1+s2=163