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# Plz help me solve this problem

Author Message
Intern
Joined: 07 Dec 2009
Posts: 8

Kudos [?]: 13 [0], given: 1

Plz help me solve this problem [#permalink]

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10 Jan 2010, 21:10
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions

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What is the easy way to solve following kind of problem?

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

a) 2^9
b) 2^10
c) 2^16
d) 2^35
e) 2^37

Kudos [?]: 13 [0], given: 1

VP
Joined: 05 Mar 2008
Posts: 1467

Kudos [?]: 307 [0], given: 31

Re: Plz help me solve this problem [#permalink]

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10 Jan 2010, 21:15
bhavinnc wrote:
What is the easy way to solve following kind of problem?

2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

a) 2^9
b) 2^10
c) 2^16
d) 2^35
e) 2^37

solve it in parts and look for a pattern

2 + 2 = 2 ^2
2^2 + 2^2 = 2^2(1+1) = 2^3
2^3 + 2^3 = 2^4

Kudos [?]: 307 [0], given: 31

Manager
Joined: 27 Apr 2008
Posts: 191

Kudos [?]: 96 [0], given: 1

Re: Plz help me solve this problem [#permalink]

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10 Jan 2010, 21:36
Or you can just use the geometric series formula.

Sum = $$\frac{a(1-r^n)}{1-r}$$

where a is the first term, r is the multiple increase, and n is the number of terms.

In this case, a=2, r=2, n=8

Solving this eq we get:
=$$\frac{2(1-2^8)}{1-2}$$
=$$\frac{2(1-256)}{-1}$$
=-2+512
=510

Now we need to add the first 2 back. Therefore:
510+2=512 = 2^9

Kudos [?]: 96 [0], given: 1

Intern
Joined: 07 Dec 2009
Posts: 8

Kudos [?]: 13 [0], given: 1

Re: Plz help me solve this problem [#permalink]

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10 Jan 2010, 22:35
Thanks, both of you,
it will help.
Regards,

Kudos [?]: 13 [0], given: 1

Re: Plz help me solve this problem   [#permalink] 10 Jan 2010, 22:35
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