Binomial Theorem,Originally Posted byfreswood

(x+y)^n = x^n +(nC1)[x^(n-1)][y] +(nC2)[x^(n-2)][y^2] +...

where

nC1 is the nCr when r=1, or is the coefficient of the 2nd term of the series/expansion.

nC2 is the nCr when r=2, or the coefficient of the 3rd term of the expansion.

nCr = [n!]/[(r!) *(n-r)!] ----------***

So,

(1 +bx)^n = 1^n +(nC1)[1^(n-1)][bx] +(nC2)[1^(n-2)][(bx)^2] +.....

First term of the expansion:

1^n is given as 1.

n cannot be solved with that because 1^n will always be 1.

For that matter, 1^(n-1) = 1; 1^(n-2) = 1; etc.....

2nd term is given as 7.5x, so,

(nC1)[1^(n-1)][bx] = 7.5x

{[n!]/[(1!) *(n-1)!]}*[1]*[bx] = 7.5x

{n}[1][bx] = 7.5x

nbx = 7.5x

The x cancels out,

nb = 7.5 --------------(1)

3rd term is given as 22.5(x^2), so,

{nC2}[1^(n-2)][(bx)^2] = 22.5(x^2)

{[n!]/[(2!)*(n-2)!]}[1][(b^2)(x^2)] = 22.5(x^2)

{[n(n-1)]/[2]}[(b^2)(x^2)] = 22.5(x^2)

The x^2 cancels out,

[(b^2)(n)(n-1)]/2 = 22.5

Multiply both sides by 2,

(b^2)(n)(n-1) = 45 --------------(2)

From (1), b = 7.5/n

Substitute that into (2),

[(7.5/n)^2](n)(n-1) = 45

(7.5/n)(7.5)(n-1) = 45

Multiply both sides by n,

(7.5)(7.5)(n-1) = 45n

56.25n -56.25 = 45n

56.25n -45n = 56.25

11.25n = 56.25

n = 56.25/11.25 = 5 ------------------answer.

And, b = 7.5/n = 7.5/5 = 1.5 ----------answer.

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I assumed you know why [n!]/[(1!) *(n-1)!] simplifies to n only.

And why [n!]/[(2!) *(n-2)!] simplifies to n(n-1)/2 only.

If you don't, you want to know why?