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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