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sequence and series.

Expert replies
by goyalsau » Mon Nov 15, 2010 3:04 am
{ 1 / (1 * 2 * 3 * 4) } + { 1 / ( 2 * 3 * 4 * 5 ) } + { 1 / 3 * 4 * 5 * 6 ) } .......... { 1 / ( 20 * 21 * 22 * 23 ) }


IN This series , The question was to find the sum of all the series..

I was able to break the first 2 terms

like 1 / 24 + 1 / 120 + 1 / 360 ..........

can be written as 1 - 23 / 24 + 23 / 24 + 110 / 120 ..............

But How to find the last fraction of this series........... If go by doing it for every term then for sure its not possible because that will be very long calculation.

Help Guys....
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Source: — Problem Solving |

by shovan85 » Mon Nov 15, 2010 11:52 am
goyalsau wrote:{ 1 / (1 * 2 * 3 * 4) } + { 1 / ( 2 * 3 * 4 * 5 ) } + { 1 / 3 * 4 * 5 * 6 ) } .......... { 1 / ( 20 * 21 * 22 * 23 ) }


IN This series , The question was to find the sum of all the series..

I was able to break the first 2 terms

like 1 / 24 + 1 / 120 + 1 / 360 ..........

can be written as 1 - 23 / 24 + 23 / 24 + 110 / 120 ..............

But How to find the last fraction of this series........... If go by doing it for every term then for sure its not possible because that will be very long calculation.

Help Guys....
How did it go? Yesterday was 14th :)

Come to problem...

{ 1 / (1 * 2 * 3 * 4) } + { 1 / ( 2 * 3 * 4 * 5 ) } + { 1 / 3 * 4 * 5 * 6 ) } .......... { 1 / ( 20 * 21 * 22 * 23 ) }
= 1/4! + 1!/5! + 2!/6! + .... + 19!/23!

[See 1 / ( 20 * 21 * 22 * 23 ) = (1*2*3*...*19)/(1*2*3*...19*20*21*22*23) = 19!/23! ]

LCM of denominator will be 23!

Now I am stuck... Anyone can try
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by Rahul@gurome » Mon Nov 15, 2010 12:39 pm
goyalsau wrote:{ 1 / (1 * 2 * 3 * 4) } + { 1 / ( 2 * 3 * 4 * 5 ) } + { 1 / 3 * 4 * 5 * 6 ) } .......... { 1 / ( 20 * 21 * 22 * 23 ) }


IN This series , The question was to find the sum of all the series..
There may be some problem with the question. The sum can be determined if the series is considered up to infinity. In case of finite summation may be we cannot determine it. Let's see what we can do with the series.

As shovan85 said, the series can be written as (0!/4! + 1!/5! + 2!/6! + 3!/7! + ... + 19!/23!)

General n-th term of the series is given by (n - 1)!/(n + 3)!
According to the question, we have to find the sum from n = 1 to n = 20.

Now, there is a summation formula for infinite factorials,
Image

If the question had asked for summation of the series up to infinity, we would've proceeded as follows,
Image

But this holds only if the summation is done for infinite series.
Rahul Lakhani
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