SOLUTION: Find the sum in terms of n for n ∑ {{{ (3k^2-5k+7) }}} k=1 Given: n ∑= {{{ (n(n+1)(2n+1))/6 }}} k=1 n ∑= {{{ (n(n+1))/2 }}} k=1

Algebra ->  Sequences-and-series -> SOLUTION: Find the sum in terms of n for n ∑ {{{ (3k^2-5k+7) }}} k=1 Given: n ∑= {{{ (n(n+1)(2n+1))/6 }}} k=1 n ∑= {{{ (n(n+1))/2 }}} k=1      Log On


   

Question 1177007: Find the sum in terms of n for
n
∑ +%283k%5E2-5k%2B7%29+
k=1
Given:
n
∑= +%28n%28n%2B1%29%282n%2B1%29%29%2F6+
k=1
n
∑= +%28n%28n%2B1%29%29%2F2+
k=1
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Your presentation of the problem is flawed; the information as given is nonsense.

I don't know if you posted it incorrectly, or if it was incorrect when it was given to you.

When the problem is presented correctly, the solution is simple.

---------------------------------------------------------

Here is how the problem should have been posted:

Find the sum in terms of n for

sum%283k%5E2-5k%2B7%2C1%2Cn%29

Given:

sum%28k%5E2%2C1%2Cn%29=%28n%28n%2B1%29%282n%2B1%29%29%2F6+ (The "k^2" was missing in your post, making the statement meaningless)

and

sum%28k%2C1%2Cn%29=%28n%28n%2B1%29%29%2F2+ (likewise, the "k" was missing)

It should also be obvious that

sum%281%2C1%2Cn%29=n

Then...



Then simplify if required.