Fool's problem of the day,

Find the sum of $ n $ terms of the series:

$$ \frac 13+ \frac{2}{21}+ \frac{3}{91} + \frac{4}{273} \cdots \text{ upto n terms } $$

Question

Fool's problem of the day,

Find the sum of $ n $ terms of the series:

$$ \frac 13+ \frac{2}{21}+ \frac{3}{91} +  \frac{4}{273} \cdots \text{ upto n terms } $$

lgbasallote · Answer

is this arithmetic progression?

anonymous · Answer

Noooway, this isn't AP.

lgbasallote · Answer

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

anonymous · Answer

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

anonymous · Answer

And please no cheating :P

lgbasallote · Answer

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

lgbasallote · Answer

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

anonymous · Answer

Yes sure you used google ? :P

lgbasallote · Answer

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

anonymous · Answer

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

callisto · Answer

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 · Answer

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 · Answer

oh wait there was...i stand corrected

callisto · Answer

anyway, i was incorrect then :D

lgbasallote · Answer

so next one is gonna be 5/ 756?

anonymous · Answer

No, the next one is not 756.

lgbasallote · Answer

651?

anonymous · Answer

HINT: The denominator is $ 1+n^2+n^4 $

lgbasallote · Answer

where n = numerator lol

lgbasallote · Answer

oh wait yeah..

anonymous · Answer

I would leave the rest to your envisage.

anonymous · Answer

How did you figure out that FoolForMath? "$1 + n^2 + n^4$"

lgbasallote · Answer

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

anonymous · Answer

Hmm, I did something like this before Ishaan ;)

mani_jha · Answer

$$n/(n ^{4}+2n ^{2}+1)-n ^{2}=n/(n ^{2}+1)^{2}-n ^{2}$$

anonymous · Answer

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

anonymous · Answer

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

mani_jha · Answer

Do we need to use the telescoping method?

anonymous · Answer

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

mani_jha · Answer

Oh well let me think..

lgbasallote · Answer

you have to brew the ingredients first

mani_jha · Answer

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 · Answer

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

mani_jha · Answer

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

mani_jha · Answer

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

anonymous · Answer

Is it 1?

mani_jha · Answer

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 · Answer

A strange answer for such a complex series :/

callisto · Answer

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

anonymous · Answer

Yeah, kind of

lgbasallote · Answer

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

callisto · Answer

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

mani_jha · Answer

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 · Answer

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

anonymous · Answer

yeahh :/

callisto · Answer

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 · Answer

you'll learn it soon :P

callisto · Answer

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

lgbasallote · Answer

there are actually cases of partial fractions

callisto · Answer

So the number of ways to do it is limited?