1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = quantity n times quantity six n squared minus three n minus one all divided by two
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

Question

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = quantity n times quantity six n squared minus three n minus one all divided by two
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

welshfella · Answer

Lets assume its trus for  n = k terms 

so 
Sk =   k (6k - 3k - 1)
          -------------
                      2

lets add the  next term (k+1)th term

S(k+1) =   k(6k - 3k - 1)    +  3((k + 1) - 2)^2
                 -----------
                          2

Now we have to try and convert this  to same form as the expression for Sk  but with  every k replaced by (k+1)

welshfella · Answer

*  thats  6k^2  in the formula  NOT 6k

welshfella · Answer

now we are aiming for 
Sk+1 =  (k+1) (6(k+1)^2  - 3(k+1) - 1)
             --------------------------
                            2


which is Sk  with (k+1) replacing k.

If we can convert the previous Sk+1 expression to this we have  in the way to the proof.

ally_b · Answer

So the statement is true for all positive integers? @welshfella

welshfella · Answer

we need to do the algebra first to see  if the  2 formula for S(k + 1) are the same . Its pretty messy I'm afraid.

ally_b · Answer

So what would I do next to solve the problem? @welshfella

welshfella · Answer

simplify the  2 expressions for  S(k+1)  and see if they are the same.

ally_b · Answer

okay thank you! @welshfella

welshfella · Answer

you'll find that they are the same
so if formula is true for n=k then its true for  n = k+1

And we know the formula is trus for  k = 1 
therfore its true for k = 2 .  3  and so on

So ny induction the formula is true for all positive integers n