# 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

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.

• $f(x)= (2x-1)^3$ is a polynomial thus there is a unique polynomial such that $g(x)-g(x-1) =f(x),g(0) =0$ so that $\sum_{k=1}^n f(k) = g(n)$ and your sum is $\sum_{k=1}^n f(k)+3f(k+1/2) = g(n)+3g(n+1/2)-3g(1/2)$ – reuns Aug 19 '19 at 3:53

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.

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: $$1^3 + 3\cdot 2^2 + 3^3 = \sum_{k=1}^{(3+1)/2}(2k-1)^3+3\sum_{k=1}^{(3+1)/2-1}(2k)^2$$

• This was my actual confusion. I got it now. Thanks. – Mathaddict Aug 19 '19 at 5:33
• You are welcome! Good luck! – farruhota Aug 19 '19 at 5:34

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 )