To find the nth power of a number

[pach2212]

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?

[wisenheimer]

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

[pach2212]

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

[arjmen]

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.

[Synergy79]

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

[arjmen]

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