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Exponents..again

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lkcr Just gettin' started! Default Avatar
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Exponents..again Post 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..Sad

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    killer1387 GMAT Destroyer! Default Avatar
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    Post 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)

    --> check the link
    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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    Post 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..Sad
    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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    Post 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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    Post 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..Sad
    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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