Question

Prove that `sum_(k=1)^n \ (2k-1)^2 = (n(2n+1)(2n-1))/3`?

Answer 1

Induction Proof - Hypothesis

We seek to prove that:

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

Which can also be written as:

`sum_(k=1)^n \ (2k-1)^2 = (n(4n^2-1))/3` ..... [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` the given result gives:

`LHS = sum_(k=1)^1 \ (2k-1)^2 = (2-1)^2 = 1`
`RHS = 1(4-1)/3 = 1`

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 \ (2k-1)^2 = (m(4m^2-1))/3` ..... [B]

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

`LHS = sum_(k=1)^(m+1) \ (2k-1)^2`
`\ \ \ \ \ \ \ \ = {sum_(k=1)^m \ (2k-1)^2 } + { (2(m+1)-1)^2 }`
`\ \ \ \ \ \ \ \ = (m(4m^2-1))/3 + (2(m+1)-1)^2 \ \ \` using [B]
`\ \ \ \ \ \ \ \ = (m(4m^2-1))/3 + (2m+2-1)^2`
`\ \ \ \ \ \ \ \ = (m(4m^2-1))/3 + (2m+1)^2`
`\ \ \ \ \ \ \ \ = 1/3{ m(4m^2-1) + 3(4m^+4m+1)}`

`\ \ \ \ \ \ \ \ = 1/3{ 4m^3-m + 12m^2+12m+3 }`

`\ \ \ \ \ \ \ \ = 1/3(m+1)(4m^2+8m+3) \ \ \` by inspection
`\ \ \ \ \ \ \ \ = 1/3(m+1)(4m^2+8m+4 - 1)`
`\ \ \ \ \ \ \ \ = 1/3(m+1)(4(m^2+2m+1) - 1)`
`\ \ \ \ \ \ \ \ = 1/3(m+1)(4(m+1)^2 - 1)`

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