anonymous
  • anonymous
Fool's problem of the day, Find the sum of \( n \) terms of the series: \[ \frac 13+ \frac{2}{21}+ \frac{3}{91} + \frac{4}{273} \cdots \text{ upto n terms } \]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lgbasallote
  • lgbasallote
is this arithmetic progression?
anonymous
  • anonymous
Noooway, this isn't AP.
lgbasallote
  • lgbasallote
fool outdid himself today :P i cant find a pattern or im just that weak hahaha

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
This took me about 10 mints to solve (without electronic aid), lets see how much you guys take :)
anonymous
  • anonymous
And please no cheating :P
lgbasallote
  • lgbasallote
it has been 8 minutes so far.....
lgbasallote
  • lgbasallote
haha finally found the pattern took me long enough :))
anonymous
  • anonymous
Yes sure you used google ? :P
lgbasallote
  • lgbasallote
no...i looked at the last term :P but i still dunno the sum
anonymous
  • anonymous
It's the pattern which is a bit elusive, the rest is kids stuff! ;)
Callisto
  • 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
  • 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
  • lgbasallote
oh wait there was...i stand corrected
Callisto
  • Callisto
anyway, i was incorrect then :D
lgbasallote
  • lgbasallote
so next one is gonna be 5/ 756?
anonymous
  • anonymous
No, the next one is not 756.
lgbasallote
  • lgbasallote
651?
anonymous
  • anonymous
HINT: The denominator is \( 1+n^2+n^4 \)
lgbasallote
  • lgbasallote
where n = numerator lol
lgbasallote
  • lgbasallote
oh wait yeah..
anonymous
  • anonymous
I would leave the rest to your envisage.
anonymous
  • anonymous
How did you figure out that FoolForMath? "\(1 + n^2 + n^4\)"
lgbasallote
  • lgbasallote
so the form is n/(n^4 + n^2 + 1)
anonymous
  • anonymous
Hmm, I did something like this before Ishaan ;)
Mani_Jha
  • Mani_Jha
\\[n/(n ^{4}+2n ^{2}+1)-n ^{2}=n/(n ^{2}+1)^{2}-n ^{2}\]
anonymous
  • anonymous
Nice problem though, most of the users would fail to see such pattern.
anonymous
  • anonymous
The problem is not finished yet. The rest is somewhat interesting :)
Mani_Jha
  • Mani_Jha
Do we need to use the telescoping method?
anonymous
  • anonymous
Yes, that's a way to do the rest, but we need to make it ready first :)
Mani_Jha
  • Mani_Jha
Oh well let me think..
lgbasallote
  • lgbasallote
you have to brew the ingredients first
Mani_Jha
  • 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
  • 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
  • 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
  • Mani_Jha
Oh, it's already done. Now just substituting n=1, n=2 etc would do
anonymous
  • anonymous
Is it 1?
Mani_Jha
  • 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
  • anonymous
A strange answer for such a complex series :/
Callisto
  • 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
  • anonymous
Yeah, kind of
lgbasallote
  • lgbasallote
looooong equaion @_@ too many numbers...fainting
Callisto
  • Callisto
no wonder why i don't understand.. thanks ..
Mani_Jha
  • 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
  • lgbasallote
haha 1 hour...it seems Fool is smarter than any of us or all of us combined :P
anonymous
  • anonymous
yeahh :/
Callisto
  • 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
  • lgbasallote
you'll learn it soon :P
Callisto
  • Callisto
After seeing similar questions for thousand times, I will :D
lgbasallote
  • lgbasallote
there are actually cases of partial fractions
Callisto
  • Callisto
So the number of ways to do it is limited?

Looking for something else?

Not the answer you are looking for? Search for more explanations.