BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

sequence and series.

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

sequence and series.

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....
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

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
If the problem is Easy Respect it, if the problem is tough Attack it
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

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
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Post new topic Post Reply
• Page 1 of 1