Chapter 12, Problem 4CT

To determine

Expansion of the sum ${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$ and evaluate the sum.

Answer to Problem 4CT

Solution:

The sum of ${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$ is $\left(\frac{2}{3}-1\right)+\left(\frac{4}{9}-2\right)+\left(\frac{8}{27}-3\right)+\left(\frac{16}{81}-4\right)=-\frac{680}{81}$.

Explanation of Solution

Given information:

${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$

Explanation:

Consider, ${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$.

As $k$ varies from $1$ to $4$,

${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}=\left[{\left(\frac{2}{3}\right)}^1-1\right]+\left[{\left(\frac{2}{3}\right)}^2-2\right]+\left[{\left(\frac{2}{3}\right)}^3-3\right]+\left[{\left(\frac{2}{3}\right)}^4-4\right]$

$=\frac{2}{3}-1+\frac{4}{9}-2+\frac{8}{27}-3+\frac{16}{81}-4$

$=\frac{130}{81}-10$

$=-\frac{680}{81}$

Thus, the sum of ${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$ is $\frac{2}{3}-1+\frac{4}{9}-2+\frac{8}{27}-3+\frac{16}{81}-4=-\frac{680}{81}$.