Q. 15 - Calculus

Question

In the proof of the Fundamental Theorem of Calculus we encounter a telescoping sum. Find the values of the following sums, which are also telescoping. 

$$\begin{gathered} \left(a\right)\sum _{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right) \\ \left(b\right) \sum _{k=1}^{100}\frac{k^2-{\left(k-1\right)}^2}{2} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a). The required sum is $\frac{100}{101}$.

Part (b). The required sum is $5000$.

Part (a) Step 1. Given Information

We are given the expressions 

$$\begin{gathered} \left(a\right)\sum _{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right) \\ \left(b\right) \sum _{k=1}^{100}\frac{k^2-{\left(k-1\right)}^2}{2} \end{gathered}$$

and we need to find their sum.

Part (a) Step 2. Explanation

The telescoping sum is:
$$\begin{gathered} \sum _{k=1}^{100}\left(\frac{1}{k}-\frac{1}{k+1}\right) \\ =-\frac{1}{100+1}+\frac{1}{1} \\ =\frac{-1+101}{101} \\ =\frac{100}{101} \end{gathered}$$

Part (b) Step 1. Explanation

The telescoping sum is:
$$\begin{gathered} \sum _{k=1}^{100}\left(\frac{k^2-{\left(k-1\right)}^2}{2}\right) \\ =\frac{{\left(1-1\right)}^2}{2}+\frac{{100}^2}{2} \\ =\frac{10000}{2} \\ =5000 \end{gathered}$$