The sum of $n$ terms of the following series; $1^3 + 3^3 + 5^3 + 7^3 + ....$ is
- $n^2(2n^2-1)$
- $n^3(n-1)$
- $n^3 + 8n + 4$
- $2n^4 + 3n^2$
The series is 1^3 + 3^3 + 5^3 + ... + (2n-1)^3. The sum of the first k cubes is (k(k+1)/2)^2. The sum of the first 2n integers is (2n(2n+1)/2)^2 = n^2(2n+1)^2. The sum of even cubes (2^3 + 4^3 + ... + (2n)^3) is 8 * (1^3 + 2^3 + ... + n^3) = 8 * (n(n+1)/2)^2 = 2n^2(n+1)^2. Subtracting even from total: n^2(2n+1)^2 - 2n^2(n+1)^2 = n^2(4n^2 + 4n + 1 - 2n^2 - 4n - 2) = n^2(2n^2 - 1).
The sum of the cubes of the first n odd natural numbers is given by the formula n squared multiplied by (2n squared - 1). You can verify this by checking the first term where n is 1, yielding 1 squared times (2 times 1 squared - 1) equals 1. Thus, the sum of n terms of the given series is n squared times (2n squared - 1).