Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms

Question

Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$

1. When $n$ is even.
2. When $n$ is odd.

This sum can be written as

$$\sum_{1}^n (2k-1)^{3} +3 \sum_{1}^n (2k)^{2} $$ I can handle the sum up to n terms when it is not specified that $n$ is even or odd. In this problem I'm confused, what changes should be done to get sum for even or odd $n$. In my textbook, $n$ is replaced by $2m$ and then they solved the problem for first $m$ terms and then substituted $m = n/2$ and same is done for odd case, by substituting $n=2m-1$. I didn't get that solution. Any suggestion would be helpful.

Answer 1

HINT

When $n = 2m$ is even, both sums have the same amount of terms, $n/2 = m$ each. When $n = 2m-1$ is odd, the left sum has one more term than the right, so there must be $m$ terms in the left and $m-1$ in the right.

Also notice that the even $n$ sum and the odd $n$ sum are different by just one last term in the right sum.

Answer 2

Details for your comment above:

$n$ is even: $$\sum_{1}^{n/2} (2k-1)^{3} +3 \sum_{1}^{n/2} (2k)^{2}$$ Example: $$1^3 + 3\cdot 2^2 = \sum_{k=1}^{2/2}(2k-1)^3+3\sum_{k=1}^{2/2}(2k)^2$$ $n$ is odd: $$\sum_{1}^{(n+1)/2} (2k-1)^{3} +3 \sum_{1}^{(n+1)/2-1} (2k)^{2}$$ Example: $$1^3 + 3\cdot 2^2 + 3^3 = \sum_{k=1}^{(3+1)/2}(2k-1)^3+3\sum_{k=1}^{(3+1)/2-1}(2k)^2$$

Answer 3

Its a hint

Split the series to two. One is sum of cubic of odd. And other one is 3*(sum of square of even )