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1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?

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1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 24 Feb 2013, 02:04
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1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?

A) 100.2^100
B) 99.2^100 + 1
C) 99.2^99 + 99
D) 2^100
[Reveal] Spoiler: OA
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 03 Mar 2013, 04:25
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shelrod007 wrote:
I still do not get the solution to this problem ??

Hi Shelrod007,

The question is: what is the value of 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99

May be I was a bit quick on the explanation. :roll:

step 1: you can rewrite this number as: 1 + 2.2^1 + 2.3.2^1 + 2.4.2^2 + 2.5.2^3 + ...... + 2.100.2^98
step 2: 1 + 2.2^1 + 2+3.2^1 + 2.4.2^2 + 2.5.2^3 + ...... + 2.100.2^98 = 1 + 2. (2^1 + 3.2^1 + 4.2^2 + 5.2^3 + ...... + 100.2^98) = 1 + 2K ==> Odd number
step 3: value of A is 100.2^100. It's an even number. Value of D is 2^100, it's an even number. So it could not be the right answer
step 4: the value of C is 99.2^99 + 99. Now, if you look into the item of the series which is 100.2^99 is equal to 99.2^99 + 2^99. compare now both values 99.2^99 + 99 < 99.2^99 + 2^99. So the value of C is smaller

hope it's clear
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 25 Feb 2013, 10:10
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process of elimination
1+ 2k is an odd number => eliminate A and D
C is smaller than 100.2^99 ==> eliminate C
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 25 Feb 2013, 09:37
is this the series like :

[0+1.2]^0 + [1+1.2]^1 + [2+1.2]^2 + [3+1.2]^3 + .....+ [99+1.2]^99

then what to do !!!!! :(
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 27 Feb 2013, 13:36
leekay wrote:
process of elimination
1+ 2k is an odd number => eliminate A and D
C is smaller than 100.2^99 ==> eliminate C


This is a great solution only if the initial question actually is:

1+ 2*2^1 + 3*2^2 + 4*2^3 + ... + 100*2^99 = ?

Is that the original question?... Or do I miss something?

Because I understood:

1+ [1+1.2]^1 + [2+1.2]^2 + [3+1.2]^3 + ... + [99+1.2]^99 = ?
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 27 Feb 2013, 21:38
johnwesley wrote:
leekay wrote:
process of elimination
1+ 2k is an odd number => eliminate A and D
C is smaller than 100.2^99 ==> eliminate C


This is a great solution only if the initial question actually is:

1+ 2*2^1 + 3*2^2 + 4*2^3 + ... + 100*2^99 = ?

Is that the original question?... Or do I miss something?

Because I understood:

1+ [1+1.2]^1 + [2+1.2]^2 + [3+1.2]^3 + ... + [99+1.2]^99 = ?


well, yeah, the initial question is 1+ 2*2^1 + 3*2^2 + 4*2^3 + ... + 100*2^99 = ? :)
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 03 Mar 2013, 03:07
I still do not get the solution to this problem ??
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 05 Apr 2013, 16:38
leekay wrote:
shelrod007 wrote:
I still do not get the solution to this problem ??

Hi Shelrod007,

The question is: what is the value of 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99

May be I was a bit quick on the explanation. :roll:

step 1: you can rewrite this number as: 1 + 2.2^1 + 2.3.2^1 + 2.4.2^2 + 2.5.2^3 + ...... + 2.100.2^98
step 2: 1 + 2.2^1 + 2+3.2^1 + 2.4.2^2 + 2.5.2^3 + ...... + 2.100.2^98 = 1 + 2. (2^1 + 3.2^1 + 4.2^2 + 5.2^3 + ...... + 100.2^98) = 1 + 2K ==> Odd number
step 3: value of A is 100.2^100. It's an even number. Value of D is 2^100, it's an even number. So it could not be the right answer
step 4: the value of C is 99.2^99 + 99. Now, if you look into the item of the series which is 100.2^99 is equal to 99.2^99 + 2^99. compare now both values 99.2^99 + 99 < 99.2^99 + 2^99. So the value of C is smaller

hope it's clear

in the above solution please let me know how did u get value of a (the first term in the series) as 100.2^100 and D as 2^100.
Also, i cant understand whats this variable C here stands for.
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 06 Apr 2013, 04:35
Though the expression seems tough to calculate, explanation by leekay makes it easy to handle.
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink] New post 06 Apr 2013, 10:18
Set I= 1+2*2^1+3*2^2+4*2^3+.....+100*2^99 then
2I= 2^1+2*2^3+3*2^3+......+99*2^99+100*2^100
- I= 1+2*2^1+3*2^2+4*2^3+.....+100*2^99
:arrow: I=- (1+2^1+2^2+2^3+......+2^99) + 100*2^100
Set G=1+2^1+2^2+2^3+......+2^99 then
2G= 2^1+2^2+2^3+....+2^100
- G=1+2^1+2^2+2^3+...+2^99
:arrow: G= 2^100 - 1
=> I= -(2^100-1) + 100*2^100 = 99*2^100 + 1
So the correct answer is B.
Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?   [#permalink] 06 Apr 2013, 10:18
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