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

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Intern
Joined: 01 Apr 2012
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1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  24 Feb 2013, 01:04
00:00

Difficulty:

45% (medium)

Question Stats:

46% (02:31) correct 54% (01:31) wrong based on 50 sessions
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
Intern
Joined: 08 Feb 2013
Posts: 6
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Kudos [?]: 3 [2] , given: 0

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  03 Mar 2013, 03:25
2
KUDOS
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.

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
Intern
Joined: 08 Feb 2013
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  25 Feb 2013, 09:10
1
KUDOS
process of elimination
1+ 2k is an odd number => eliminate A and D
C is smaller than 100.2^99 ==> eliminate C
Intern
Joined: 05 Jun 2012
Posts: 31
GMAT 1: 480 Q48 V9
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Kudos [?]: 2 [0], given: 18

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  25 Feb 2013, 08:37
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 !!!!!
Manager
Joined: 24 Jan 2013
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Kudos [?]: 48 [0], given: 6

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  27 Feb 2013, 12:36
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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Intern
Joined: 08 Feb 2013
Posts: 6
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Kudos [?]: 3 [0], given: 0

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  27 Feb 2013, 20:38
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 = ?
Intern
Joined: 23 Jan 2013
Posts: 38
Concentration: Technology, Other
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Kudos [?]: 0 [0], given: 9

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  03 Mar 2013, 02:07
I still do not get the solution to this problem ??
Intern
Joined: 11 Feb 2013
Posts: 20
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Kudos [?]: 2 [0], given: 4

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  05 Apr 2013, 15:38
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.

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.
Senior Manager
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  06 Apr 2013, 03:35
Though the expression seems tough to calculate, explanation by leekay makes it easy to handle.
Intern
Joined: 06 Jan 2013
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Kudos [?]: 0 [0], given: 12

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  06 Apr 2013, 09:18
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
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
G= 2^100 - 1
=> I= -(2^100-1) + 100*2^100 = 99*2^100 + 1
So the correct answer is B.
Manager
Joined: 24 Nov 2012
Posts: 150
Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44
WE: Business Development (Internet and New Media)
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Kudos [?]: 34 [0], given: 69

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  09 Aug 2013, 06:23
mihir66 wrote:
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

Took some time for me to figure it out but here is my attempt...

the nth term of the series is n.2^(n-1) and the first few terms of the series is 1, 4, 12, 32 -------(1)

The answers can be broken down as

A = n*2^n
B = [(n-1) * 2^n] + 1
C = 99.2^99 + 99 = 99(2^99+1) = (n-1)[(2^n-1) +1]
D = 2^n

Substitute n = 2, Answer should be 5 as per (1)

The correct answer is got only by B
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Senior Manager
Joined: 10 Jul 2013
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Kudos [?]: 68 [0], given: 102

Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ? [#permalink]  09 Aug 2013, 12:30
mihir66 wrote:
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

.......................

30 seconds solution:

1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = An odd number, because 1+ even(everything is a multiple of 2).
so the Answer must be an odd number.
so A and D eliminated.
Now the last term of the series is 100.2^99 which is just 98 greater than option C) 99.2^99 + 99. (subtract and see).
but this is a series of 100 terms and even the 5th term itself is 80 . so its not possible for option (C), ultimately we have 1 option and its (B)
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Re: 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?   [#permalink] 09 Aug 2013, 12:30
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# 1+ 2.2^1 + 3.2^2 + 4.2^3 + ... + 100.2^99 = ?

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