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sequence

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Re: sequence

by lightbulb » Wed Oct 29, 2008 7:39 pm
Hi, here's my take on it:

1/(1*2) = 1 - 1/2
1/(2*3) = 1/2 - 1/3
1/(3*4) = 1/3 - 1/4 and so on

So, we have:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)....

Irrespective of whether n is even or odd, you have:
S = (1 - 1/2) + (1/2 - 1/3) + .... + (1/n - 1/(n+1))
= 1 - 1/(n+1) ( notice how the terms cancel each other)
= n/(n+1)

earth@work wrote:how do we find the sum of n terms of series : (1/(1*2)) + (1/(2*3)) + (1/(3*4)) + .........
(not a multiple choice question)
Answer : [spoiler]n/(n+1)[/spoiler]
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by earth@work » Wed Oct 29, 2008 7:53 pm
Thanks a lot lightbulb! nice short method !
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Re: sequence

by logitech » Wed Oct 29, 2008 7:55 pm
lightbulb wrote:Hi, here's my take on it:

1/(1*2) = 1 - 1/2
1/(2*3) = 1/2 - 1/3
1/(3*4) = 1/3 - 1/4 and so on

So, we have:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5)....

Irrespective of whether n is even or odd, you have:
S = (1 - 1/2) + (1/2 - 1/3) + .... + (1/n - 1/(n+1))
= 1 - 1/(n+1) ( notice how the terms cancel each other)
= n/(n+1)

earth@work wrote:how do we find the sum of n terms of series : (1/(1*2)) + (1/(2*3)) + (1/(3*4)) + .........
(not a multiple choice question)
Answer : [spoiler]n/(n+1)[/spoiler]
Great work Lightbulb! 8)

earth@work:

1/[nx(n+1)] = 1/n - 1/(n+1)

This will help you out as Lightbulb described.
LGTCH
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by earth@work » Wed Oct 29, 2008 7:57 pm
thanks for the formula logitech, it will be very useful.
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