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Thread: To find the nth power of a number

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    Within my grasp! pach2212's Avatar
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    To find the nth power of a number

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    Guys,

    Is there any easy way to find out the nth power of a number.

    Suppose a problem is formulated this way:

    What is the total money you get back if you invest $100 compounded half yearly at 8% pa for 6 years.

    The value will be = 100(1.04)^12

    How do I go about from here?

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    Use the binomial theorem :

    (1+x)^n= 1+nx +n(n-1)/2!*x^2+n(n-1)(n-2)/3!*x^3 .......

    For the above problem, you can do this till 2 steps to get an approximate answer.
    100(1+.04)^12= 100(1+.04*12 + (.04^2)*12*11/2 .........)
    = 158.56

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    Thanks wisenheimer.

    I am aware of this binomial theorem.

    I din't use it since the value of the power (i.e. 12 ) was too high. Was looking for any other method if exists.

    However the answer is 160.10

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    Grrr... arjmen's Avatar
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    I can give you an approximate value -

    (1.04)^12 = [(((1+0.04)^2)^2)^2]*((1+0.4)^2)^2

    (1+0.04)^2 = 1+0.08+0.0016 ~ 1.08
    (1+0.08)^2 = 1+0.16+0.0064 ~ 1.17
    (1+0.17)^2 = 1+0.34+0.0289 ~ 1.37

    (1.04)^12 ~ (1+0.37)*(1+17) = 1+0.37+0.17+0.0629 ~ 1.60

    100*(1.04)^12 ~ 160

    But the more accuracy you need, the longer the time taken to calculate.

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    Guys,

    I remember from my school days that we had specific formulae to calculate sum of the series such as

    1+2+3+4+....+n
    1^2+2^2+3^2+....+n^2
    1^3+2^3+3^3+...+n^3

    Does somebody has these formulae and if so can he/she please post them.

    Thanks

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    Quote Originally Posted by Synergy79
    I remember from my school days that we had specific formulae to calculate sum of the series such as
    1+2+3+4+....+n
    1^2+2^2+3^2+....+n^2
    1^3+2^3+3^3+...+n^3
    http://www.ilovemaths.com/3sequence.htm

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