Suppose that the first 3 terms of the binomial expansion of 1(+ax^3)^n are 1, -18x^3, 144x^6.Find the values?

Question

Suppose that the first 3 terms of the binomial expansion of 1(+ax^3)^n are 1, -18x^3, 144x^6.Find the values of a and n

Answer 1

Hi,

The binomial expansion of (1 + ax^3)^n would start with these 3 terms:

1^n + n*1^(n-1)*(ax^3)^1 + [n(n-1)/2]*1^(n - 2)*(ax^3)^2

This simplifies to:

1 + n*ax^3 + [n(n-1)/2]*a^2*x^6

Since the second term is -18x^3, then -18x^3 = n*ax^3, so
-18 = n*a.

Since the third term is 144x^6, then [n(n-1)/2]*a^2*x^6 = 144x^6, so [n(n-1)/2]*a^2 = 144.

Our system of 2 equations to find "a" and "n" is:

-18 = n*a
[n(n-1)/2]*a^2 = 144

Solving the first equation for "a", it becomes: a = -18/n
Substituting this into the second equation, it becomes:

n(n - 1)* (-18)^2
---------------------- = 144 Simpliying the fraction:
__2_____(n)^2

n(n - 1)* (324)
---------------------- = 144 Cancelling and reducing:
__2n^2

(n - 1)* (162)
---------------------- = 144 Multiplying by n:
__n

(n - 1)* (162) = 144n
162n - 162 = 144n
18n = 162
n = 9

If n = 9, then substituting 9 for n in -18 = n*a, it becomes:

-18 = 9*a Dividing by 9,

-2 = a

So a = -2 and n = 9

I hope that helps.

Answer 2

i anticipate you could doe the 1st area. Then replace x via (x-x^2) giving a million+5(x-x^2)+10(x-x^2)^2+10(x-x^2)^3 and boost the brackets and carry jointly like words ignoring powers larger than x^3. I make it a million+5x+5x^2-10x^3 As a examine, utilising your calculator sub (say) x=0.a million into (a million+x-x^2)^5 giving a million.53862 and right into a million+5x+5x^2-10x^3 giving a million.fifty 4 that's on the edge of the almost good answer. This confirms your working.

Answer 3

a=2 and n=9
since na=18 and a^2(n^2-n)=288