Chapter F, Problem 18E

To determine

To write: The sum $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ in sigma notation.

Answer to Problem 18E

The sum $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ expressed in sigma notation as ${\displaystyle \sum _{i=1}^6\frac{1}{i^2}}$.

Explanation of Solution

Definition used:

If $a_m,a_{m+1},\mathrm{...},a_n$ are real numbers and m and n are integers such that $m\le n$, then ${\displaystyle \sum _{i=m}^n{a}_i=a_m+}{a}_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_n$.

Calculation:

Here, the sum $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ is the sum of 6 numbers and number is of the form $2^i$, where $i$ varies from 0 to 5.

Thus, by the above definition the sum $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ expressed as follows.

$\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}={\displaystyle \sum _{i=1}^6\frac{1}{i^2}}$

Thus, the sum $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ expressed in sigma notation as ${\displaystyle \sum _{i=1}^6\frac{1}{i^2}}$.