Original question: Use binomial theorem to find #(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
