Asked by menaga

expand (x+8y)^5 using the binomial theorem up to x^3. hence find the value of (1.08)^5 correct two decimal place.
Answered by Steve
(x+8y)^5 = x^5 + 5x^4(8y) + 10x^3(8y)^2 + ...
= x^5 + 50x^4y + 640x^3y^2 + ...

Now just plug in x=1 and y=.01
Answered by Damon
if only allowed up to x^3
do it in two steps
result for (x+8y)^3 * (x+8y)^2
that is easier anyway