Question

S = 1^2 + 3^2 + 5^2 + ... + (2n-1)^2... sum of squared odd numbers. a clue please?

Answer 1

Consider the possibilities:

If n is even.
If n is odd.

Also remember the square of an odd is always odd.

Answer 2

i knew that for 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

Answer 3

Do you know sum of squares = n(n+1)(2n+1)/6

Answer 4

Let T(n)= n(n+1)(2n+1)/6
then S(n) can be expressed as a combination of T(n)

Answer 5

I've got an idea. Let n=2k,
\[N(k)=1^2+2^2+..+(2k)^2=\frac{2k(2k+1)(4k+1)}{6}.\]
The sum of squared odd numbers between 1 and 2k is the sum of all squared numbers minus the sum of the squared even numbers \(E(k)\),
\[E(k)=2^2+4^2+..(2k)^2=2^2(1^2+2^2+..+k^2)=\frac{4k(k+1)(2k+1)}{6}.\]
And hence \(S=N-E\).

Answer 6

If you do the subtraction you get \(S=\frac{1}{3}k(4k^2-1)\). Substitute n=2k and that's it.