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

Asked by Pragya Singh | 1 year ago |  167

1 Answer

Solution :-

0.99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)5 = (1 – 0.01)5

5C(1)5 – 5C(1)4 (0.01) + 5C(1)3 (0.01)2

= 1 – 5 (0.01) + 10 (0.01)2

= 1 – 0.05 + 0.001

= 0.951

Answered by Abhisek | 1 year ago

Related Questions

Find the term independent of x in the expansion of (3/2 x2 – 1/3x)9

Find the term independent of x in the expansion of $$(\dfrac{3}{2x^2} – \dfrac{1}{3x})^9$$

Class 11 Maths Binomial Theorem View Answer

Find the middle term in the expansion of (x – 1/x)2n+1

Find the middle term in the expansion of $$(x-\dfrac{ 1}{x})^{2n+1}$$

Class 11 Maths Binomial Theorem View Answer

Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Class 11 Maths Binomial Theorem View Answer

Find the middle term in the expansion of (1 – 2x + x2)n

Find the middle term in the expansion of (1 – 2x + x2)n

Class 11 Maths Binomial Theorem View Answer

Find the middle term in the expansion of (x/a – a/x)10

Find the middle term in the expansion of $$(\dfrac{x}{a} – \dfrac{a}{x})^{10}$$

Class 11 Maths Binomial Theorem View Answer