binomial theorem, with two unknowns

[freswood]

Even my teacher can't figure this one out!

The first 3 terms in the expansion of (1+bx)^n are

1 + 7.5x + 22.5x^2

find b and n.

The answer: b = 1.5, n = 5.

By the way, this is done using the binomial theorem. Thanks everyone for your help!

[krtxmrtz]

Using this notation:

{n,m}=n! / [m! (n-m)!]

and using the binomial theorem:

(1 + bx)ⁿ = {n,0} + {n,1}bx + {n,2}b²x² + ... = 1 + nbx + n(n-1)b²x²/2 + ...

And identifying terms:

nb = 7.5
n(n-1)b²/2 = 22.5

From the first of these two:

b = 7.5 / n

and substituting this one into the other:

n(n-1)(7.5 / n)²/2 = 22.5

and this leads to

(n-1) / n = 45 / 7.5²

n = 1 / [1 - (45 / 7.5²)] = 5

b = 7.5 / n = 1.5

[freswood]

Thanks so much! That makes perfect sense now.

[freswood]

I just found another one I can't figure out:

In the exapansion of (2a - 1)^n, the coefficient of the second term is -192. Find the value of n.

I can't seem to do it because it only gives one coefficient.

This is as far as I got (I hope it's right): 2^(n-1) = 192/n

[krtxmrtz]

Indeed, if only one coefficient is given then n can have more than one value. Actually the expression you get is:

(2a)^n-1 = 192/n

For example, n = 2 is a solution, as long as a = 43. Or n = 3 and a = 4 is another solution. Or n = 6 and a = 1. But not n = 5 for then you'd have 192/n on the right hand side, which is not integer.

[Guv]

The first answer seems incorrect to me. Check the following

(1 + bx)ⁿ = 1ⁿ + b*1^n-1*x + (n-1)*b*1^n-2*x²

Compare the above with 1 + 7.5x + 22.5x² + additional terms

From the above, b = 7.5 and b(n-1) = 22.5 or 22.5 = 7.5(n-1)


The above gives (n-1) = 3 or n = 4

[krtxmrtz]

...Check the following

(1 + bx)ⁿ = 1ⁿ + b*1^n-1*x + (n-1)*b*1^n-2*x²

The expansion is nearly correct, but:

(1 + bx)ⁿ = 1ⁿ + b*1^n-1*x + (n-1)*b²*1^n-2*x²