How do you use the binomial theorem to approximate (1.02)^8(1.02)8?

1 Answer
May 12, 2017

See below.

Explanation:

Original question: Use binomial theorem to find (1.02)^8(1.02)8

Consider the binomial theorem:
(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b^1+...+((n),(n-1))a^1b^n-1+((n),(n))a^0b^n
where n is a positive integer and ((x),(y)) is x choose y.

Since the binomial theorem only works on values in the form of a binomial: Consider that 1.02=1+0.02=1+1/50

So, by substituting 1.02=1+1/50, we get:
(1+1/50)^8
By applying the binomial theorem, we get:
=((8),(0))+((8),(1))(1/50)+((8),(2))(1/50)^2+((8),(3))(1/50)^3+((8),(4))(1/50)^4+((8),(5))(1/50)^5+((8),(6))(1/50)^6+((8),(7))(1/50)^7+((8),(8))(1/50)^8
=1+8/50+28/50^2+56/50^3+70/50^4+56/50^5+28/50^6+8/50^7+1/50^8
=45767944570401/39062600000000~~1.17166 rounded to 5 decimal places

We can confirm this result:
(1.02)^8~~1.17166 rounded to 5 decimal places