7.Binomial Theorem
medium

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

A

$0.951$

B

$0.951$

C

$0.951$

D

$0.951$

Solution

$0.99=1-0.01$

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

$ = {\,^5}{C_0}{(1)^5} – {\,^5}{C_1}{(1)^4}(0.01) + {\,^5}{C_2}{(1)^3}{(0.01)^2}$        [ Approximately ]

$=1-5(0.01)+10(0.01)^{2}$

$=1-0.05+0.001$

$=1.001-0.05$

$=0.951$

Thus, the value of $(0.99)^{5}$ is approximately $0.951$

Standard 11
Mathematics

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