Find an approximation of `(0. 99)^5`using the first three terms of its expansion.

Explanation

We know that

`(a+b)^n=^nC_0a^n+^nC_1a^(n-1)b^1+^nC_2a^(n-2)b^2+.......+^nC_(n-1)a^1b^(n-1)+^nC_nb^n`

Hence,
`(a+b)^5=^5C_0a^5+^5C_1a^4b^1+^5C_2a^3b^2+^5C_3a^2b^3+^5C_4a^1b^4+^5C_5b^5`

`(a+b)^5=(5!)/(0!(5-0)!)a^5+(5!)/(1!(5-1)!)a^4b^1+(5!)/(2!(5-2)!)a^3b^2+(5!)/(3!(5-3)!)a^2b^3+(5!)/(4!(5-4)!)a^1b^4+(5!)/(5!(5-5)!)b^5`

`(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5a^1b^4+b^5`

`(0.99)^5=(1-0.01)^5`

Using First three terms from `(a+b)^5`

`(0.99)^5=1-0.05+0.001`

`(0.99)^5=1.001-0.050`

`(0.99)^5=0.9510`

So, approximate value of `(0.99)^5= 0.9510`