Q. 38 - Calculus

Question

Evaluate the finite sums.

$\sum _{k=1}^{50} \frac{1}{3^k}$

Step-by-Step Solution

Answer

The sum of the series $\sum _{k=1}^{50} \frac{1}{3^k}$ is, $\frac{1-{\left({\displaystyle \frac{1}{3}}\right)}^{50}}{2}$.

Step 1 . Given Information

$\sum _{k=1}^{50} \frac{1}{3^k}$.

Step 2 . Let us expand the given series.

$\sum _{k=1}^{50} \frac{1}{3^k}=\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{50}}$

The first term in the series $a=\frac{1}{3}$

The common ratio is, $\frac{1}{3}$ with the number of terms, $n=50$.

Step 3 . The finite sum of the geometric series with ratio less than 1 is given by,

$S_n=\frac{a\left(1-r^n\right)}{1-r}$

Substitute the known values in the formula.

$$\begin{gathered} S_{50}=\frac{{\displaystyle \frac{1}{3}}\left(1-{\left({\displaystyle \frac{1}{3}}\right)}^{50}\right)}{1-{\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \frac{1}{3}}-{\left({\displaystyle \frac{1}{3}}\right)}^{51}}{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ =\frac{1-{\left({\displaystyle \frac{1}{3}}\right)}^{50}}{2} \end{gathered}$$