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## Exponents..again

This topic has 2 expert replies and 2 member replies
lkcr Just gettin' started!
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Exponents..again Sun Apr 15, 2012 2:29 am
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• Lap #[LAPCOUNT] ([LAPTIME])
1. 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =?

Why does that equal 2^9? Am I missing something.....
(^ = to the power of)

2. (1001^2-999^2) /(101^2-99^2)

What would be the best approach to handle 2.?

Sorry for these silly questions..

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killer1387 GMAT Destroyer!
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Sun Apr 15, 2012 4:46 am
1. 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =?

Why does that equal 2^9? Am I missing something.....
(^ = to the power of)

http://www.beatthegmat.com/2-2-2-2-2-3-2-4-2-5-2-6-2-7-2-8-t109134.html

2. (1001^2-999^2) /(101^2-99^2)

--> use a^2-b^2= (a+b)(a-b)

(1001^2-999^2) /(101^2-99^2)
= 2000*2/200*2
=10

hope this helps..!!

Shalabh's Quants Really wants to Beat The GMAT!
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Sun Apr 15, 2012 10:15 pm
lkcr wrote:
1. 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =?

Why does that equal 2^9? Am I missing something.....
(^ = to the power of)

2. (1001^2-999^2) /(101^2-99^2)

What would be the best approach to handle 2.?

Sorry for these silly questions..
Its Geometric progression question ...

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

=2+(2^1+2^2+.......+2^7+2^8)

=2+[2*(2^8-1)]/(2-1)=2+2^9-2; Sum of GP = a.(r^n - 1)/(r-1); where a = I term = 2, r = common ratio = 2, n = no. of terms

=2^9=512

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Shalabh Jain,
e-GMAT Instructor

Last edited by Shalabh's Quants on Sun Apr 15, 2012 10:46 pm; edited 1 time in total

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:32 pm
lkcr wrote:
1. 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =?

Why does that equal 2^9? Am I missing something.....
(^ = to the power of)
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8
= 2 + 2(1 + 2 + 2² + ... + 2^7)
Now, 1 + 2 + 2² + ... + 2^7 is a geometric series, where first term = 2 and the common ratio = 2 > 1

Geometric Series is of the form: a + ar + ar² + ... + ar^(n - 1)
Here, a = first term
r = common ratio
n = number of terms
Then sum of n terms of geometric series = a(r^n - 1)/(r - 1) when r > 1 and a(1 - r^n)/(1 - r) when r < 1

In the given case, 1 + 2 + 2² + ... + 2^7 = 1(2^8 - 1)/(2 - 1) = 2^8 - 1

Therefore, 2 + 2(1 + 2 + 2² + ... + 2^7) = 2 + 2(2^8 - 1) = 2 + 2^9 - 2 = 2^9 = 512

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### GMAT/MBA Expert

Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:36 pm
lkcr wrote:
2. (1001^2-999^2) /(101^2-99^2)

What would be the best approach to handle 2.?

Sorry for these silly questions..
We know that a² - b² = (a - b)(a + b), so apply this identity in the given question.
(1001² - 999²)/(101² - 99²) = (1001 - 999)(1001 + 999)/(101 - 99)(101 + 99)
= (2)(2000)/(2)(200)
= 2000/200
= 10

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Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

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