Question

Find an approximation of $(0.99{)}^5$ using the first three of its expansion.

• A. $0.95$
• B. $0.97$
• C. $0.98$
• D. $89$

Answer 1

The correct option is A $0.95$

$(0.99{)}^5=(1-0.01{)}^5$

We know that

$(a+b{)}^n{=}^n{C}_0{a}^n{+}^n{C}_1{a}^{n-1}{b}^1{+}^n{C}_2{a}^{n-2}{b}^2+-----{+}^n{C}_{n-1}{a}^1{b}^{n-1}{+}^n{C}_n{b}^n$

Hence

$(a+b{)}^5{=}^5{C}_0{a}^5+{}^5{C}_1{a}^4{b}^1+{}^5{C}_2{a}^3{b}^2{+}^5{C}_3{a}^2{b}^2{+}^5{C}_{4}a{b}^4+{}^5{C}_5{b}^5$

$=a^5+\frac{5!}{1!(5-1)!}{a}^4{b}^1+\frac{5!}{2!(5-2)!}{a}^3{b}^2+\frac{5!}{3!(5-3)!}{a}^2{b}^3+\frac{5!}{4!(5-4)!}a{b}^4+b^5$

Using first three terms,

$(0.99{)}^5=1-0.05+0.001$

$=1.001-0.050$

$=0.9510$

So, approximate value of $(0.99{)}^5=0.9510$