Home
Class 12
MATHS
Find an approximation of (0. 99)^5 using...

Find an approximation of `(0. 99)^5` using the first three terms of its expansion.

Text Solution

Verified by Experts

`0.99 = 1 - 0.01`
`:. (0.999)^(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`.
Promotional Banner

Topper's Solved these Questions

  • BINOMIAL THEOREM

    CENGAGE|Exercise Exercise 8.1|17 Videos

  • BINOMIAL THEOREM

    CENGAGE|Exercise Exercise 8.2|10 Videos

  • AREA UNDER CURVES

    CENGAGE|Exercise Question Bank|20 Videos

  • BINOMIAL THEORM

    CENGAGE|Exercise Question Bank|31 Videos

Similar Questions

Explore conceptually related problems

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

(iii) Find an approximate value of (0.99)^(5) using the first three terms of its expansion.

Write the first three terms of the expansion (1+q^(2))^(10).

Find the a,b, and n in th expansion of (a+b)^(n) if the first three terms of the expansion are 729,7290, and 30375, respectively.

Find a,b and n in the expansion of (a+b)^(n) if the first three terms of the expansion are 729,7290 and 30375, respectively.

If the coefficients of the three successive terms in the binomial expansion of (1+x)^(n) are in the ratio 1:4.42 then the first of these terms in the expansion is

If the sum of coefficient of first half terms in the expansion of (x+y)^(n) is 256, then find the greatest coefficient in the expansion.