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
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?
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.
http://www.ilovemaths.com/3sequence.htmOriginally Posted by Synergy79
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks