Respuesta :
Given: The below
[tex]\sum_{n\mathop{=}1}^{\infty}10(-\frac{3}{8})^n[/tex]To Determine the sum to infinity of the given
Solution
Letus re-write the give as below
[tex]\sum_{n\mathop{=}1}^{\infty}10(-\frac{3}{8})^n=10\sum_{n\mathop{=}1}^{\infty}(-\frac{3}{8})^n[/tex]The sum is give by the formula below
[tex]S=\frac{a}{1-r}[/tex]From the given, we can found that a and r is as shown below
[tex]\begin{gathered} a=-\frac{3}{8} \\ r=-\frac{3}{8} \end{gathered}[/tex]Therefore
[tex]\begin{gathered} S=\frac{a}{1-r} \\ 1-r=1-(-\frac{3}{8}) \\ =1+\frac{3}{8} \\ =\frac{8+3}{8} \\ =\frac{11}{8} \\ S=\frac{-\frac{3}{8}}{\frac{11}{8}} \end{gathered}[/tex][tex]\begin{gathered} S=-\frac{3}{8}\div\frac{11}{8} \\ S=-\frac{3}{8}\times\frac{8}{11} \\ S=-\frac{3}{11} \end{gathered}[/tex]Substituting the sum as shown below
[tex]\begin{gathered} S=\sum_{n\mathop{=}1}^{\infty}(-\frac{3}{8})^n=-\frac{3}{11} \\ Therefore \\ 10\sum_{n\mathop{=}1}^{\infty}(-\frac{3}{8})^n=10\times-\frac{3}{11}=-\frac{30}{11} \end{gathered}[/tex]Hence, the equivalent to sum to infinity of the given is -30/11, OPTION C
