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

Answer:

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.