Understanding Closed Forms of Series | Quick Explanation and Tips

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Hi, I'm just going through an example of closed forms of series in a textbook (Anton Calc) and I'm wondering how they got to this step, can someone please explain it to me? Also, any hints on formulating closed forms would be greatly appreciated, they are not covered in detail in my text. Thanks a lot!

Homework Statement



\sum_{k=o}^{3+n}{k^2} - 14<br /> =\frac{(3+n)(4+n)(7+2n)}{6} - 14

Sorry if it's a stupid question, but I've never tried working with a sum with variable at the top, and I don't know what I'm doing.

Homework Equations


Given above

The Attempt at a Solution


None, this is an example problem I don't understand.

Edit - there should be brackets surrounding the sum that cut off the -14 as it's not part of the sum. I couldn't work this out in LaTeX and I'm in a hurry.
 
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iamaelephant said:
Hi, I'm just going through an example of closed forms of series in a textbook (Anton Calc) and I'm wondering how they got to this step, can someone please explain it to me? Also, any hints on formulating closed forms would be greatly appreciated, they are not covered in detail in my text. Thanks a lot!

Homework Statement



\sum_{k=o}^{3+n}{k^2} - 14<br /> =\frac{(3+n)(4+n)(7+2n)}{6} - 14

Sorry if it's a stupid question, but I've never tried working with a sum with variable at the top, and I don't know what I'm doing.

Homework Equations


Given above


The Attempt at a Solution


None, this is an example problem I don't understand.

Edit - there should be brackets surrounding the sum that cut off the -14 as it's not part of the sum. I couldn't work this out in LaTeX and I'm in a hurry.

the first part is the sum of first x squares (in this case x = 3+n)
\sum_{k=o}^{x}{k^2} = \frac{x(x+1)(2x+1)}{6} plug in x= 3+n then u get
\frac{(3+n)(4+n)(7+2n)}{6}
of course u can prove these sums to the first "x" terms of k^n (in ur case it is k^2)..
 
Thanks a lot! Heh, always seems so simple when someone explains it :)
 
Is anyone able to give me some hints or links on how to find the closed form of any series? I'm really stuck here and my textbooks don't seem to cover it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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