Chapter 12, Problem 5CT

To determine

The sum $-\frac{2}{5}+\frac{3}{6}-\frac{4}{7}+\mathrm{...}+\frac{11}{14}$ using summation notation.

Answer to Problem 5CT

Solution:

The sum $-\frac{2}{5}+\frac{3}{6}-\frac{4}{7}+\mathrm{...}+\frac{11}{14}$ using summation notation is ${\displaystyle \sum _{k=1}^{10}{(-1)}^k}\left(\frac{k+1}{k+4}\right)$.

Explanation of Solution

Given information:

The sum, $-\frac{2}{5}+\frac{3}{6}-\frac{4}{7}+\mathrm{...}+\frac{11}{14}$.

Explanation:

Consider, the sum $-\frac{2}{5}+\frac{3}{6}-\frac{4}{7}+\mathrm{...}+\frac{11}{14}$.

Observe that, the signs of the terms are alternate negative and positive. This implies that there will be a general term with power of $-1$.

Also note, numerator is 1 more than the term number and denominator is 4 more than the tem number.

Thus, we can write each term in general as:

${(-1)}^k\left(\frac{k+1}{k+4}\right)$

The last term of the numerator is 11, which tells that there are 10 terms.

Thus the required sum in summation notation is:

$-\frac{2}{5}+\frac{3}{6}-\frac{4}{7}+\mathrm{...}+\frac{11}{14}={\displaystyle \sum _{k=1}^{10}{(-1)}^k}\left(\frac{k+1}{k+4}\right)$.