Question

How to write the statement `S_1,S_2,S_3`? Thanks!

Answer 1

We have:

`S_n: 1^2 + 4^2 + 7^2 + ... (3n-2)^2 = (n(6n^2-3n-1))/2`

So then statements `S_1, S_2 and S_3` would be:

`S_1: 1^2 = (1(6-3-1))/2`
`\ \ \ \ \ \ => 1 = 2/2 = 1`, which is true .

`S_2: 1^2 +4^2= (2(24-6-1))/2`
`\ \ \ \ \ \ => 1+16 = 18-1`, which is also true .

`S_3: 1^2 +4^2 + 7^2= (3(54-9-1))/2`
`\ \ \ \ \ \ => 1+16 +49 = 66`, which is also true .

In general, we can prove the result is true using Induction:

Induction Proof - Hypothesis

We seek to prove that:

`sum_(k=1)^n \ (3k-2)^2 = (n(6n^2-3n-1))/2` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` we have `S_1`, which we have already shown to be true/

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`sum_(k=1)^m \ (3k-2)^2 = (m(6m^2-3m-1))/2` ..... [B]

Consider the LHS of [A] with the addition of the next term, in which case we have

`LHS = sum_(k=1)^(m+1) \ (3k-2)^2`
`\ \ \ \ \ \ \ \ = {sum_(k=1)^m \ (3k-2)^2 } + (3(m+1)-2)^2`
`\ \ \ \ \ \ \ \ = (m(6m^2-3m-1))/2 + (3m+1)^2 \ \ \` using [B]
`\ \ \ \ \ \ \ \ = (m(6m^2-3m-1) + 2(3m+1)^2)/3`
`\ \ \ \ \ \ \ \ = (6m^3-3m^2-m + 2(9m^2+6m+1))/2`
`\ \ \ \ \ \ \ \ = (6m^3-3m^2-m + 18m^2+12m+2)/2`
`\ \ \ \ \ \ \ \ = (6m^3+ 15m^2+11m+2)/2`
`\ \ \ \ \ \ \ \ = ((m+1)(6m^2+9m+2))/2`
`\ \ \ \ \ \ \ \ = ((m+1)(6(m^2+2m+1)-12m-6+9m+2))/2`
`\ \ \ \ \ \ \ \ = ((m+1)(6(m+1)^2-3m-4))/2`
`\ \ \ \ \ \ \ \ = ((m+1)(6(m+1)^2-3(m+1)-1))/2`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` (ie, `S_m => S_(m+1)`) where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`sum_(k=1)^n \ (3k-2)^2 = (n(6n^2-3n-1))/2 \ \ \ QED`