Chapter F, Problem 41E

Evaluate each telescoping sum.

1. (a) ${\displaystyle \underset{i=1}{\overset{n}{\sum }}[i^4-{(i-1)}^4]}$
2. (b) ${\displaystyle \underset{i=1}{\overset{100}{\sum }}(5^i-5^{i-1})}$
3. (c) ${\displaystyle \underset{i=3}{\overset{99}{\sum }}\left(\frac{1}{i}-\frac{1}{i+1}\right)}$
4. (d) ${\displaystyle \underset{i=1}{\overset{n}{\sum }}(a_i-a_{i-1})}$

To determine

To find: The value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left[i^4-{\left(i-1\right)}^4\right]}$.

Answer to Problem 41E

The value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left[i^4-{\left(i-1\right)}^4\right]}$ is $n^4$.

Explanation of Solution

Simplify the expression ${\displaystyle \sum _{i=1}^n\left[i^4-{\left(i-1\right)}^4\right]}$ and obtain the value of the sum.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left[i^4-{\left(i-1\right)}^4\right]}=\left(1^4-0^4\right)+\left(2^4-1^4\right)+\left(3^4-2^4\right)+\cdots +\left(n^4-{\left(n-1\right)}^4\right)\\ =n^4-0^4\\ =n^4\end{array}$

Thus, the value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left[i^4-{\left(i-1\right)}^4\right]}$ is $n^4$.

To determine

To find: The value of the telescoping sum ${\displaystyle \sum _{i=1}^{100}\left[5^i-5^{i-1}\right]}$.

Answer to Problem 41E

The value of the telescoping sum ${\displaystyle \sum _{i=1}^{100}\left[5^i-5^{i-1}\right]}$ is $5^{100}-1$.

Explanation of Solution

Simplify the expression ${\displaystyle \sum _{i=1}^{100}\left[5^i-5^{i-1}\right]}$ and obtain the value of the sum.

$\begin{array}{c}{\displaystyle \sum _{i=1}^{100}\left[5^i-5^{i-1}\right]}=\left(5^1-5^{1-1}\right)+\left(5^2-5^{2-1}\right)+\left(5^3-5^{3-1}\right)+\cdots +\left(5^{100}-5^{100-1}\right)\\ =\left(5^1-5^0\right)+\left(5^2-5^1\right)+\left(5^3-5^2\right)+\cdots +\left(5^{100}-5^{99}\right)\\ =5^{100}-5^0\\ =5^{100}-1\end{array}$

Thus, the value of the telescoping sum ${\displaystyle \sum _{i=1}^{100}\left[5^i-5^{i-1}\right]}$ is $5^{100}-1$.

To determine

To find: The value of the telescoping sum ${\displaystyle \sum _{i=3}^{99}\left[\frac{1}{i}-\frac{1}{i+1}\right]}$.

Answer to Problem 41E

The value of the telescoping sum ${\displaystyle \sum _{i=3}^{99}\left[\frac{1}{i}-\frac{1}{i+1}\right]}$ is $\frac{97}{100}$.

Explanation of Solution

Simplify the expression ${\displaystyle \sum _{i=3}^{99}\left[\frac{1}{i}-\frac{1}{i+1}\right]}$ and obtain the value of the sum.

$\begin{array}{c}{\displaystyle \sum _{i=3}^{99}\left[\frac{1}{i}-\frac{1}{i+1}\right]}=\left(\frac{1}{3}-\frac{1}{3+1}\right)+\left(\frac{1}{4}-\frac{1}{4+1}\right)+\left(\frac{1}{5}-\frac{1}{5+1}\right)+\cdots +\left(\frac{1}{99}-\frac{1}{99+1}\right)\\ =\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots +\left(\frac{1}{99}-\frac{1}{100}\right)\\ =\frac{1}{3}-\frac{1}{100}\\ =\frac{97}{100}\end{array}$

Thus, the value of the telescoping sum ${\displaystyle \sum _{i=3}^{99}\left[\frac{1}{i}-\frac{1}{i+1}\right]}$ is $\frac{97}{100}$.

To determine

To find: The value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left(a_i-a_{i-1}\right)}$.

Answer to Problem 41E

The value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left(a_i-a_{i-1}\right)}$ is $a_n-a_0$.

Explanation of Solution

Simplify the expression ${\displaystyle \sum _{i=3}^n\left(a_i-a_{i-1}\right)}$ and obtain the value of the sum.

$\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(a_i-a_{i-1}\right)}=\left(a_1-a_{1-1}\right)+\left(a_2-a_{2-1}\right)+\left(a_3-a_{3-1}\right)+\cdots +\left(a_n-a_{n-1}\right)\\ =\left(a_1-a_0\right)+\left(a_2-a_1\right)+\left(a_3-a_2\right)+\cdots +\left(a_n-a_{n-1}\right)\\ =a_n-a_0\end{array}$

Thus, the value of the telescoping sum ${\displaystyle \sum _{i=1}^n\left(a_i-a_{i-1}\right)}$ is $a_n-a_0$.