Question

Expand (1-3x)^8 in ascending power of x up to x³.hence using our expansion evaluate 0.97^8 correct to 5 decimal places.?

Answer 1

`(0.97)^8~=0.78374`

Explanation:

Using Pascal's triangle

Expansion of `(1-3x)^8` is

`1-8*3x+28*(3x)^2-56*(3x)^3+70*(3x)^4-.................`

=0`1-24x+252x^2- 1512x^3+5670x^4-.................`

Now if `x=0.01` then `0.97^8=(1-3xx0.01)^8`

= `1-24(0.01)+252(0.01)^2- 1512(0.01)^3+5670(0.01)^4-.................`

= `1-0.24+0.0252-0.001512+0.0005670-...........`

`~=0.78374`