# a) Find the first 4 terms in ascending powers of x of (1-x/4)^10 b) By using a suitable substitution, use your result in (a) to approximate (0.975)^10

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### 1 Answer

a) You need to use binomial expansion such that:

`(x+y)^n = C_n^0*x^n + C_n^1*x^(n-1)*y + C_n^2*x^(n-2)*y^2 + ... + C_n^n y^n`

Substituting 1 for x, `-x/4` for `y` and `10` for`n` yields:

`(1 - x/4)^10 = C_10^0*1^10- C_10^1*1^(10-1)*(x/4) + C_10^2*1^(10-2)*(x/4)^2- C_10^3*1^(10-3)*(x/4)^3... + C_10^10*(x/4)^10`

You need to evaluate binomial coefficients such that:

`C_10^0 = 1`

`C_10^1 = 10`

`C_10^2 = (10!)/(2!(10-2)!) =gt C_10^2 = (8!9*10)/(2!(8)!)`

`C_10^2 = 9*10/2 = 45`

`C_10^3 = (10!)/(3!(10-3)!) =gtC_10^3 = (7!8*9*10)/(3!(7)!)`

`C_10^3 = 120`

**Hence, the first four terms of binomial expansion, in ascending powers of x are `1 , -10*(x/4) , 45*(x/4)^2 , -120*(x/4)^3.` **b) You need to set `(1-x/4)^10` equal to `(0.975)^10` such that:

`(1-x/4)^10 = (0.975)^10 =gt 1 - x/4 = 0.975 =gt x/4 = 1 - 0.975 =gt x/4 = 0.025 =gt x = 0.1`

**Hence,using binomial `(1-x/4)^10 ` and the given value `(0.975)^10` yields `x=0.1` **