1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\frac{1}{13}-\frac{1}{16}.............
not a doubt[4]
hint-think DI
21@ankur well wats teh need to do all tat.....and btw ans is not 0
the given series is nothing but this integral \int_{0}^{1}{\frac{dx}{1+x^{3}}} as pointed out by saumya and eure
\int_{0}^{1}{\frac{dx}{1+x^{3}}}=\int_{0}^{1}{\frac{dx}{(1+x)(1-x+x^2)}}
now resolve it into partial fractions\int_{0}^{1}{\left[\frac{1}{3(1+x)}-\frac{(x-2)}{3(1+x^2-x) } \right]}dx
now it can be done
the ans which i got was \frac{1}{3}\left[log2+\frac{\pi }{\sqrt{3}} \right]
1@ Soumya:
Haan... i got the method of how you formulated your series.... thts nice..
But what is wrong with my method? plz do tell me...
and how did eureka123 get his formula?
1Irs very easy , as you see that the denominators are in ap with common diff of 3 , so if the powers
of x are also in ap with common diff 3 starting from 1 , then we are done , as the powers come in
the denominators if you integrate . and the limits only decide the numerators , so they are easy to
determine too .
1@ Soumya:
But how does the integration of that thing give you the sum of that series? Plz give me the steps on how you arrived at that integral from the given sequence.
1arrey ankur , 1 / 1 + x^3 is not the general term , it is just what we want to integrate --- in order
to get the sum -- plzz read posts 5 and 6.
Thnx eragon 24 , beautifully done .
21well as far as how it is......jus integrate the infinite gp series with limits from 0 to 1 given by saumya.......u will get teh req series
1I got the general term using the following concept :
The series is a harmonic progression because the consecutive denominators are in A.P.
Hence general term is :
11 + (n-1)3
This gives 13n - 2
1@eragon
Well, I told you I'm not sure about the answer.
Btw, how did you get the general term as 11 + x3 ?
Cuz none of the terms seem to fit that general formula.
1the sum s = 3 / 4 + 1 /7 - 1/10 + 1/ 13 - 1/16
gen term excluding 3/4 , t = (-1)n-1 14n - n + 4
= ( -1 )n 13n + 4
would this help ??????
1We split the series into odd and even terms.
First we find the sum of the series given by :
1 + 14 + 110.... up to 'n' terms.
that is,
0∫n 13n-2 dn
From this we subtract twice the sum of even terms. Sum of even terms is given by :
0∫n/2 13(2n) - 2 dn
So the series becomes :
0∫n 13n-2 dn - 2 0∫n/2 13(2n) - 2 dn
Integrating this, we get:
13 [ loge (3n-2) ]0n - 26 [ loge (6n-2) ]0n/2
Simplifying, this expression we get :
13 [ loge (3n-2) - loge (-2) ] - 13 [ loge (3n-2) - loge (-2) ]
Subtract both, and we get answer as 0
I don't know if the answer is correct, though....
But I have applied correct basic integration to the question.
1Club the two terms as :
(1+ 1/7 + 1/13 +...) - (1/4 + 1/10 + 1/16 + .........)
Now apply some approximation method.
Draw the graphs of these two curves and find the area of the curves.
Subtract the 2nd from the first.
I hope u will get there.
24u r rite soumya...and it is clearly a GP
0∫111-(-x3)
=> 0∫111+x3
1Is the answer \int_{0}^{1}{} [ 1 - x3 + x6 -x9 + ......] dx ?
Well , I haven't still learned enough integration to compute this one :( : (
1well u both hav just found out teh general term..
think in terms of definite integration[1]
