Question

Evaluate the finite sums.

$\sum _{k=0}^{100} \frac{1}{2^k}$

Short answer

The sum of the series$\sum _{k=0}^{100} \frac{1}{2^k}$is,$2-\frac{1}{2^{100}}$.

Step-by-step solution

Step 1. Given Information

$\sum _{k=0}^{100} \frac{1}{2^k}$.

Step 2. Let us expand the given series.

$\sum _{k=0}^{100} \frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{100}}$.

The first term in the series, $a=1$.

The common ratio is, $r=\frac{1}{2}$with number of terms$n=101$.

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.

$$\begin{gathered} S_{101}=\frac{1\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^{101}\right)}{1-{\displaystyle \frac{1}{2}}} \\ =\frac{1-{\displaystyle \frac{1}{2^{101}}}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ =2\left(1-\frac{1}{2^{101}}\right) \\ =2-\frac{1}{2^{100}} \end{gathered}$$