Prove the statement by mathematical induction?

If S_n=1+2xx2^2+3^2+2xx4^2+5^2+2xx6^2 then S_n=n(n+1)^2/2,n is even
S_n=n^2(n+1)/2,n is odd

1 Answer
May 22, 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(n):1+2xx2^2+3^2+...+2xx(n-1)^2+n^2=n^2(n+1)/2

if n is odd and

tau_2(n):1+2xx2^2+3^2+...+(n-1)^2+2xxn^2=n(n+1)^2/2

if n is even.

Let's prove tau_1, firstly. The base case, tau_1(1), claims that

1=1^2(1+1)/2 -= "True".

Now, let x be an odd integer such that tau_1(x) is true.

:. color(red)(1+2xx2^2+3^2+...+2xx(x-1)^2+x^2=x^2(x+1)/2)=S_x

We must prove if the next case holds. However, the next case is not tau_1(x+1). This is becaused we defined tau_1 only on odd positive integers. This means that the next case is actually tau_1(x+2), as x+2 is odd.

tau_1(x+2):color(red)(1+2xx2^2+...+2xx(x-1)^2+x^2)+2xx(x+1)^2+(x+2)^2=(x+2)^2(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(x+1)/2

So by substituting this into the statement, we get

color(red)(x^2(x+1)/2)+2xx(x+1)^2+(x+2)^2 = (x+2)^2(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+7x^2+16x+12 = x^3+7x^2+16x+12

Which is true. Thus, by mathematical induction, tau_1(n) is true for all odd integers n.

The second statement, tau_2, is similar. Check the base case tau_2(2), 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(n) is true for n even.

Hence, by Mathematical induction, the statement

S_n :

{(1+2xx2^2+...+2xx(n-1)^2+n^2=n^2(n+1)/2", if n is odd"),(1+2xx2^2+...(n-1)^2+2xxn^2 = n(n+1)^2/2", if n is even"):}

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