Question

Prove by mathematical induction. 1 + 5 + 9 +...+ (4n-3) = (2n+1)(n-1) for n ≥ 2

Answer 1

Let $P(n)$ be the proposition that

$$1 + 5 + 9 +...+ (4n-3) =n(2n-1)$$

for $n\geq1.$

Basis Step

$P(1)$ is true, because $1=1(2(1)-1).$

Inductive Step

For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k.$ That is, we assume that

$$1 + 5 + 9 +...+ (4k-3) =k(2k-1)$$

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

$$1 + 5 + 9 +...+ (4k-3)+(4(k+1)-3)$$

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

is also true.

When we add $(4(k + 1)-3)$ to both sides of the equation in $P(k),$ we obtain

$$1 + 5 + 9 +...+ (4k-3)+(4(k+1)-3)$$

$$=k(2k-1)+(4(k + 1)-3)$$

$$=2k^2-k+4k+1$$

$$=2k^2+3k+1$$

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

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

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 proven that

$$1 + 5 + 9 +...+ (4n-3) =n(2n-1)$$

for $n\geq1.$