Fool's problem of the day,
Find the sum of \( n \) terms of the series:
\[ \frac 13+ \frac{2}{21}+ \frac{3}{91} + \frac{4}{273} \cdots \text{ upto n terms } \]

- anonymous

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- chestercat

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- lgbasallote

is this arithmetic progression?

- anonymous

Noooway, this isn't AP.

- lgbasallote

fool outdid himself today :P i cant find a pattern or im just that weak hahaha

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## More answers

- anonymous

This took me about 10 mints to solve (without electronic aid), lets see how much you guys take :)

- anonymous

And please no cheating :P

- lgbasallote

it has been 8 minutes so far.....

- lgbasallote

haha finally found the pattern took me long enough :))

- anonymous

Yes sure you used google ? :P

- lgbasallote

no...i looked at the last term :P but i still dunno the sum

- anonymous

It's the pattern which is a bit elusive, the rest is kids stuff! ;)

- Callisto

the denominator is like, 1x3, 3x7, 7x13, 13x21 and the numerator is like the difference of the 2 numbers divided by 2.. Nah.. but don't even know what to do..

- lgbasallote

there's no pattern in the denominator..it is inconsistent in callisto's interpretation i think...the pairs are 1-3, 3-7, 7-13, 13-21...there is no pattern on the second digit of the pair

- lgbasallote

oh wait there was...i stand corrected

- Callisto

anyway, i was incorrect then :D

- lgbasallote

so next one is gonna be 5/ 756?

- anonymous

No, the next one is not 756.

- lgbasallote

651?

- anonymous

HINT: The denominator is \( 1+n^2+n^4 \)

- lgbasallote

where n = numerator lol

- lgbasallote

oh wait yeah..

- anonymous

I would leave the rest to your envisage.

- anonymous

How did you figure out that FoolForMath? "\(1 + n^2 + n^4\)"

- lgbasallote

so the form is n/(n^4 + n^2 + 1)

- anonymous

Hmm, I did something like this before Ishaan ;)

- Mani_Jha

\\[n/(n ^{4}+2n ^{2}+1)-n ^{2}=n/(n ^{2}+1)^{2}-n ^{2}\]

- anonymous

Nice problem though, most of the users would fail to see such pattern.

- anonymous

The problem is not finished yet. The rest is somewhat interesting :)

- Mani_Jha

Do we need to use the telescoping method?

- anonymous

Yes, that's a way to do the rest, but we need to make it ready first :)

- Mani_Jha

Oh well let me think..

- lgbasallote

you have to brew the ingredients first

- Mani_Jha

We've to express the general term(nth term) as a difference of the nth and (n+1)th terms, or nth and (n-1)th terms. Come on guys and gals, think!

- anonymous

\[\frac{n}{(n^2 + 1 - n)(n^2+1+n)} = \frac 1 2\left(\frac{1}{n^2+1 - n} - \frac{1}{n^2 +1+n}\right)\]

- Mani_Jha

\[\sum_{1}^{n}n/(n^4+n^2+1)=\sum_{1}^{n}(n^2+n+1)-(n^2-n+1)/2(n^2+n+1)(n^2-n+1)\]
=\[\sum_{1}^{n}((1/n^2-n+1)-(1/n^2+n+1))/2\]

- Mani_Jha

Oh, it's already done. Now just substituting n=1, n=2 etc would do

- anonymous

Is it 1?

- Mani_Jha

Yes, it seems so. If we put n=1 we get 1 for the first term and the other terms keep getting cancelled on summation.

- anonymous

A strange answer for such a complex series :/

- Callisto

\[\frac{n}{(n^2 + 1 - n)(n^2+1+n)} = \frac 1 2\left(\frac{1}{n^2+1 - n} - \frac{1}{n^2 +1+n}\right)\]
Is it something related to partial fraction?

- anonymous

Yeah, kind of

- lgbasallote

looooong equaion @_@ too many numbers...fainting

- Callisto

no wonder why i don't understand.. thanks ..

- Mani_Jha

That's not so difficult, Callisto. It is similar to the following operation:
\[1/n(n+1)=(n+1)-n/n(n+1)\]
=\[(n+1)/n(n+1)-n/n(n+1)=1/n-1/(n+1)\]
If you can understand the above, you would've no problem understanding this:
He just wrote
\[n=2n/2={(n^2+n+1)-(n^2-n+1)}/2\]
Then each term above was divided by the denominator. Ok?

- lgbasallote

haha 1 hour...it seems Fool is smarter than any of us or all of us combined :P

- anonymous

yeahh :/

- Callisto

but sometimes it's difficult to think how to break down the fraction, my teacher hasn't taught us anything about partial fraction except
1/[n(n+1)] = 1/n - 1/(n+1).. (just remembering)
So, whenever i see something related to partial fraction, probably i'll give up at this stage :S

- lgbasallote

you'll learn it soon :P

- Callisto

After seeing similar questions for thousand times, I will :D

- lgbasallote

there are actually cases of partial fractions

- Callisto

So the number of ways to do it is limited?

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