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  4. The approximation of (0.99)5 using the first three terms of its expansion is

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Q. The approximation of $(0.99)^5$ using the first three terms of its expansion is

Binomial Theorem

Solution:

Now, $(0.99)^5=(1-0.01)^5$
$={ }^5 C_0(1)^5-{ }^5 C_1(1)^4(0.01)+{ }^5 C_2(1)^3(0.01)^2$
(ignore the other terms)
$ =1-5 \times 1 \times 0.01+\frac{5 \times 4}{2} \times 1 \times 0.01 \times 0.01 $
$ =1-0.05+10 \times 0.0001=1-0.05+0.001 $
$=1.001-0.05=0.951$