1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f(x)= (2x-1)^3$ 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 $\sum_{k=1}^n f(k)+3f(k+1/2) = g(n)+3g(n+1/2)-3g(1/2)$ $\endgroup$ – reuns Aug 19 '19 at 3:53
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This was my actual confusion. I got it now. Thanks. $\endgroup$ – Mathaddict Aug 19 '19 at 5:33
  • $\begingroup$ You are welcome! Good luck! $\endgroup$ – farruhota Aug 19 '19 at 5:34
0
$\begingroup$

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 )

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.