Answer to Question #251523 in Discrete Mathematics for Gayathri

Question

The sum of the first n positive odd integers is n².

Establish the formula for the sum of the first n positive even integers and use proof by mathematical induction to prove its correctness.

Answer 1

The sum of the first $n$ positive even integers is $n(n+1).$

Let $P(n)$ be the proposition that the sum of the first n positive even integers is $n(n+1).$

Basic Step: $P(1)$ is true because $2=1(1+1).$

Inductive Step: For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary

positive integer $k.$ That is, we assume that

$$2+4+...+2k=k(k+1)$$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$$2+4+...+2k+2(k+1)=(k+1)(k+1+1)$$

is also true.

When we add $2(k + 1)$ to both sides of the equation in $P(k),$ we obtain

$$2+4+...+2k+2(k+1)=k(k+1)+2(k+1)$$

$$=(k+1)(k+2)$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers n. That is, we have proved that

$$2+4+...+2n=n(n+1)$$

for all positive integers n.

The sum of the first $n$ positive even integers is $n(n+1).$