Q. 41 - Calculus

Question

Evaluate the finite sums.

$\sum _{k=6}^{150} {\left(\frac{2}{3}\right)}^k$

Step-by-Step Solution

Answer

On evaluating, we get,

$3\left({\left(\frac{2}{3}\right)}^6-{\left(\frac{2}{3}\right)}^{151}\right)$

Step 1. Given information.

Consider the given question,

$\sum _{k=6}^{150} {\left(\frac{2}{3}\right)}^k$

Step 2. Use the sum of n terms of geometric series.

The geometric series $\sum _{k=6}^{150} {\left(\frac{2}{3}\right)}^k$ in the expanded form is $\sum _{k=1}^{150}{\left(\frac{2}{3}\right)}^k={\left(\frac{2}{3}\right)}^6+{\left(\frac{2}{3}\right)}^7+...+{\left(\frac{2}{3}\right)}^{150}$.

The series has the first term $a={\left(\frac{2}{3}\right)}^6$.

The common ratio $r=\frac{2}{3}$ with no. of terms,

$n=145$

The finite sum of the geometric series with ratio greater than $1$, $S_n=\frac{a\left(r^n-1\right)}{r-1}$.

Substituting the values,

$$\begin{gathered} S_{145}=\frac{{\left({\displaystyle \frac{2}{3}}\right)}^6\left(1-{\left(\frac{2}{3}\right)}^{145}\right)}{1-\left(\frac{2}{3}\right)} \\ =\frac{3{\left({\displaystyle \frac{2}{3}}\right)}^6\left(1-{\left(\frac{2}{3}\right)}^{145}\right)}{3-2} \\ =3\left({\left(\frac{2}{3}\right)}^6-{\left(\frac{2}{3}\right)}^{151}\right) \end{gathered}$$