# ?

?

## 16 Answers

FIRST CALCULATE THE Tnth TERM

IT IS

n/((2n+1)!)

just sum it out to find answer

sigma(n/(2n+1)!)

sigma(.5*2n/(2n+1)!)

i.e

.5*sigma((2n+1-1)/(2n+1)!)

.5*sigma(1/(2n-1)!+1/(2n+1)!)

.5*(1/1!-1/3!+1/3!-1/5!..........-1/(2n+1)!)

cancelling we get

0.5*(1-1/(2n+1)!)

that is the answer

the question is not toughest as you say

For me,its tough!Im a year younger than u.....Anyways thnks for ur ans

Hi Satyaram,

That is because, nobody gets the r-th term of the series right in this.

The r-th teerm in the series is

T_{r} = r(r!)2^{r}/(2r+1)!.

And this is not quite easy to sum. You try to sum this.

Dont ask for the solution. It is a good summation question to do on your own.

Regards,

Ashwin (IIT Madras).

Dude,

The difficult thing is to find the rth term...

How did it strike to u that T_{r} = r(r!)2^{r}/(2r+1)! when u saw the question?

Dude,

When you look at some particular term, say for example the 4th term which is

4/(1.3.5.7.9)

the obvious thing is the Nr is "r"

now comes the tricky part, the Dr (the number of terms in the Dr varies, and are products of odd numbers)

So you have 1.3.5.7.9 - now to represent this in mathematicl notation you need to have (1.2.3.4.5.6.7.8.9)

So 1.3.5.7.9 = (1.2.3.4.5.6.7.8.9)/(2.4.6.8) = (9!)/2^{4}4!

so for the r-th term the Dr is (2r+1)!/2^{r}r!

And hence you get the Tr.

Regards,

Ashwin (IIT Madras).

But then that is not the toughest part.

The tougher part is to sum it. :-)

Regards,

Ashwin (IIT Madras).

Dude,

I got an more easier solution dude :

Multiply and divide by 2.

Now,

Write the numerator as difference of the first and last term.i.e.

Write

and

and so on....

Now take the 1/2 common and:::

u get

which is

Hey this is what was my answer.

I admit my answer is wrong but what is wrong with my nth term

Harikrishnan,

The n-th term is not

n/(2n+1)!. For eg put n=2, your expression will give 2/5! = 2/(1.2.3.4.5). But in the question the 2nd term is 2/(1.3.5).

Ok Satyaram,

For of all there is no solution to your question anywhere above, for you to give a simpler solution :-)

And more over the solution has not been typed!!!.... it has been copy-pasted :-)

That solution is the only way to solve it.

Nice that you got it.

Regards,

Ashwin (IIT Madras).

Dude,

Macha,I got the answer from another friend.

This is latex dude :) and he finished it in easier way...

And Ive seen this type of question,actually its an trignometric series,in which after simplyifying it...they use this method,this method is common for all the series question i think so :)

And Of course,i copy-pasted it .... (Ennaala Eppidi yosikamudiyum boss.........)

And also I want to know ur solution ..........

And also see artofproblemsolving.com ,be an member and its also an common forum for all maths olympiads all over the world,MIT students r the experts of this site (Like IItians in askiitian ),u can post,ask and clear ur doubts :)

Dude,

That is the only solution, as I had stated previously.

Do the same thing with the r-th term [ie the general term of the series, so that it will be over with just one term]

If you note

Tr = r(r!)2^{r}/(2r+1)!

= 1/2 [ (r-1)!2^{r-1}/(2r-1)! - r!2^{r}/(2r+1)! ]

= 1/2 [ t_{r-1} - t_{r} ]

So, the summation upto infinite terms = 1/2

And summation upto n-terms is 1/2 [ 1 - n!2^{n}/(2n+1)! ]

Regards,

Ashwin (IIT Madras).

omg i am sorry by (2n+1)! i mean you have to substitue 1 then 2 and so on to n and not substituting n and calulating the factorial.

ok now it is okkk is nt it

or if confusion persists

just replace (2n+1)!(at all places where you see it) with ∏(2n+1) where ∏ denotes product just like sigma giving n values from 1 to n

then the final answer would be

.5(1-1/(∏(2n+1)))

or

.5(1-(2^n)*n!/(2n+1)!)

hope you all are clear with it

sorry for any misconceptions

Harikrishnan,

Exactly. Now the change is fine.

Though your method was right, your final answer was wrong just because you had mis-interpreted the n-th term of the series.

Regards,

Ashwin (IIT Madras).