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# Sum

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Manager
Joined: 25 Mar 2009
Posts: 54

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08 Apr 2009, 01:19
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S=1/2+1/6+1/12..............+1/10100, Find S.
Intern
Joined: 07 Feb 2009
Posts: 47

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08 Apr 2009, 05:20
Nice question Mikko,

I enjoyed solving this.

Explanation:

S=1/2+1/6+1/12..............+1/10100
or, S=(1-1/2)+(1/2-1/3)+(1/3-1/4)+..............+(1/100-1/101),
every term gets canceled except first and last.

so, S=1-1/101
or, S=100/101.
Manager
Joined: 02 Mar 2009
Posts: 130

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08 Apr 2009, 08:38
Fantastic Abhishek! How did you think of this? i could hardly spot the pattern!

Thanks
Manager
Joined: 25 Mar 2009
Posts: 54

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09 Apr 2009, 03:48
abhishekik wrote:
Nice question Mikko,

I enjoyed solving this.

Explanation:

S=1/2+1/6+1/12..............+1/10100
or, S=(1-1/2)+(1/2-1/3)+(1/3-1/4)+..............+(1/100-1/101),
every term gets canceled except first and last.

so, S=1-1/101
or, S=100/101.

Yes, totaly agree with you, a little more detailed:

1/2=1/(1*2)=(2-1)/2*3=1/2-1/3 (it 's more obvious when u put it in fraction )
1/6=1/(2*3)=1/2-1/3
.....
Director
Joined: 01 Apr 2008
Posts: 847
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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10 Apr 2009, 01:39
Dude how did u think abt this pattern...can you tell the line of thought? I tried to see if it is a AP or GP..but then couldnt think of this pattern...
abhishekik wrote:
Nice question Mikko,

I enjoyed solving this.

Explanation:

S=1/2+1/6+1/12..............+1/10100
or, S=(1-1/2)+(1/2-1/3)+(1/3-1/4)+..............+(1/100-1/101),
every term gets canceled except first and last.

so, S=1-1/101
or, S=100/101.
Intern
Joined: 07 Feb 2009
Posts: 47

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10 Apr 2009, 05:55
10100 helped me to find this pattern.
10100 clearly looked like 101x100, then I came to know that 2 is also 2x1, 6- 3x2, 12- 4x3 etc.
and I just tried the ways to break 1/2 using 2x1, 1/6 using 3x2 and so on. Finally, I got this solution.
Re: Sum   [#permalink] 10 Apr 2009, 05:55
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