Question

Prove that${\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}{\mathbf{5}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{\left)}\mathbf{\right(}\mathbf{2}\mathit{n}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{/}\mathbf{3}$ whenever nis a nonnegative integer

Short answer

It is proved that$1^2+3^2+5^2+\dots +(2n+1{)}^2=(n+1\left)\right(2n+1\left)\right(2n+3)/3$ whenever nis a nonnegative integer.

Step-by-step solution

Principle of Mathematical Induction

To prove that is true for all positive integers n, whereis a propositional function, we complete two steps:

Basis Step:

We verify that P(1)is true.

Inductive Step:

We show that the conditional statement $\mathit{P}\left(K\right)\mathbf{\to }\mathit{P}\left(k+1\right)$is true for all positive integers k.

Proving the basis step

Given statement is

$1^2+3^2+5^2+\dots +(2n+1{)}^2=\frac{(n+1)(2n+1)(2n+3)}{3}$

In the basis step, we need to prove that P(1) is true

For finding statement P(1) substituting 1 for n in the statement

Therefore, the statement P(1) is

$\begin{array}{r}{1}^2+\left(2\right(1)+1{)}^2=\frac{(1+1)\left(2\right(1)+1)\left(2\right(1)+3)}{3}\\ 10=\frac{30}{3}\\ 10=10\end{array}$

Therefore, the statement P(1) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, we need to prove that, if P(k) is true, then P(K+1) is also true.

That is,

$P\left(k\right)\to P\left(K+1\right)$is true for all positive integers k.

In the inductive hypothesis, we assume that is true for any arbitrary positive integer

That is

$1^2+3^2+5^2+\dots +(2k+1{)}^2=\frac{(k+1)(2k+1)(2k+3)}{3}$...(i)

Now we must have to show that P ( K + 1 ) is also true

Therefore replacing K with K + 1 in the statement

$\begin{array}{r}{1}^2+3^2+5^2+\dots +\left(2\right(k+1)+1{)}^2=\frac{\left(\right(k+1)+1)\left(2\right(k+1)+1)\left(2\right(k+1)+3)}{3}\\ =\frac{(k+2)(2k+3)(2k+5)}{3}\\ =\frac{\left(2k^2+3k+4k+6\right)(2k+5)}{3}\\ =\frac{\left(2k^2+7k+6\right)(2k+5)}{3}\\ =\frac{4k^3+14k^2+12k+10k^2+35k+30}{3}\\ =\frac{4k^3+24k^2+47k+30}{3}\end{array}$

Now, Adding ${\left(2\left(K+1\right)+1\right)}^2$in both sides of the equation (i) or inductive hypothesis.