Chapter 4, Problem 6R

Explanation of Solution

Sum of all even numbers from 0 to 2n:

If the sum is even number of integers that is “$\text{E}\left(\text{n}\right)$”, then it is clear that the sum of all even numbers is “$\text{2S}\left(\text{n}\right)$”.

Here, “$\text{S}\left(\text{n}\right)$” denotes that sum of all integers from “1” to “n” and the sum of all even integers is “$\text{n}\left(\text{n+1}\right)$”.

Proof:

Let us prove the result by following induction,

For base case “$\text{n=1}$ ”,

• Assume that “$\text{Sum }\left(\text{2×i}\right)$” for the value $\text{i=0…n}$ is “$\text{n}\left(\text{n+1}\right)$”
• It is the same for “$\text{n+1}$”.
• “$\text{Sum}\left(\text{2}\times \text{i}\right)$” for $\text{i=0…n+1=}\left(\text{n+1}\right)\left(\left(\text{n+1}\right)\text{+1}\right)\text{=}\left(\text{n+1}\right)\left(\text{n+2}\right){\text{ = n}}^{\text{2}}\text{+3n+2}$