Prove the statement by mathematical induction?

If ${S}_{n} = 1 + 2 \times {2}^{2} + {3}^{2} + 2 \times {4}^{2} + {5}^{2} + 2 \times {6}^{2}$ then ${S}_{n} = n {\left(n + 1\right)}^{2} / 2 , n$ $i s$ even ${S}_{n} = {n}^{2} \frac{n + 1}{2} , n$ $i s$ odd

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Hammer Share
May 24, 2018

See below.

Explanation:

Since we have two different situations, let's break down this into two different statements, ${\tau}_{1}$ and ${\tau}_{2}$, respectively.

${\tau}_{1} \left(n\right) : 1 + 2 \times {2}^{2} + {3}^{2} + \ldots + 2 \times {\left(n - 1\right)}^{2} + {n}^{2} = {n}^{2} \frac{n + 1}{2}$

if $n$ is odd and

${\tau}_{2} \left(n\right) : 1 + 2 \times {2}^{2} + {3}^{2} + \ldots + {\left(n - 1\right)}^{2} + 2 \times {n}^{2} = n {\left(n + 1\right)}^{2} / 2$

if $n$ is even.

Let's prove ${\tau}_{1}$, firstly. The base case, ${\tau}_{1} \left(1\right)$, claims that

$1 = {1}^{2} \frac{1 + 1}{2} \equiv \text{True}$.

Now, let $x$ be an odd integer such that ${\tau}_{1} \left(x\right)$ is true.

$\therefore \textcolor{red}{1 + 2 \times {2}^{2} + {3}^{2} + \ldots + 2 \times {\left(x - 1\right)}^{2} + {x}^{2} = {x}^{2} \frac{x + 1}{2}} = {S}_{x}$

We must prove if the next case holds. However, the next case is not ${\tau}_{1} \left(x + 1\right)$. This is becaused we defined ${\tau}_{1}$ only on odd positive integers. This means that the next case is actually ${\tau}_{1} \left(x + 2\right)$, as $x + 2$ is odd.

${\tau}_{1} \left(x + 2\right) : \textcolor{red}{1 + 2 \times {2}^{2} + \ldots + 2 \times {\left(x - 1\right)}^{2} + {x}^{2}} + 2 \times {\left(x + 1\right)}^{2} + {\left(x + 2\right)}^{2} = {\left(x + 2\right)}^{2} \frac{x + 3}{2} = {S}_{x + 2}$

Notice how the part highlighted in red is the former sum, ${S}_{x}$.
We know that

${S}_{x} = {x}^{2} \frac{x + 1}{2}$

So by substituting this into the statement, we get

$\textcolor{red}{{x}^{2} \frac{x + 1}{2}} + 2 \times {\left(x + 1\right)}^{2} + {\left(x + 2\right)}^{2} = {\left(x + 2\right)}^{2} \frac{x + 3}{2}$

This is fairly easy to find. Multiply both sides by $2$, expand all squares and multiply the parantheses (or, in other words, do it directly) to get the following result:

${x}^{3} + 7 {x}^{2} + 16 x + 12 = {x}^{3} + 7 {x}^{2} + 16 x + 12$

Which is true. Thus, by mathematical induction, ${\tau}_{1} \left(n\right)$ is true for all odd integers $n$.

The second statement, ${\tau}_{2}$, is similar. Check the base case ${\tau}_{2} \left(2\right)$, assume it is true for some even whole number $y$ then check it for the next even number, $y + 2$.
If you do all that, you will find that ${\tau}_{2} \left(n\right)$ is true for $n$ even.

Hence, by Mathematical induction, the statement

${S}_{n} :$

$\left\{\begin{matrix}1 + 2 \times {2}^{2} + \ldots + 2 \times {\left(n - 1\right)}^{2} + {n}^{2} = {n}^{2} \frac{n + 1}{2} \text{ & if n is odd" \\ 1+2xx2^2+...(n-1)^2+2xxn^2 = n(n+1)^2/2" & if n is even}\end{matrix}\right.$

is true for all $n$ who follow all the respective conditions.

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