Um, just had a little problem the other day during a test.

The first three terms in the expansion of the binomial (a+b)^n are 729, 2916 and 4860 respectively. Find a, b and n.

I started with

(a+b)^n = a^n + ^{n}{C}_{1}(a)^{n-1}(b)^1 + ^{n}{C}_{2}(a)^{n-2}(b)^2

Then equated the terms;

a^n = 729
n(a)^{n-1}(b) = 2916
\frac{n(n-1)}{2}(a)^{n-2}(b)^2 = 4860

I then tried to substitute log_a 729 in the other equations but I seemed to get nowhere.

I got frustrated and started on another trail; trial and error. Then I got a=3, b=2 and n=6. I know these are the answers but I don't know how to get them the proper way. Please help!

Hey there:

Using the three equations you have, I get a=3, b=2, n=6.

I solved the first one for n and got n=\frac{6ln(3)}{ln(a)}

Then, I just used substitution for the others and got:

b=\frac{2aln(a)}{3ln(3)}

This gives n=6, a=3, b=2

I would have done it like this - it's not as elegant as Galactus's approach, but may be a bit easier to follow:

You know that the first coefficient is a^n = 729. So consider what are the square roots, cube roots, etc of 729 that are integers? If you convert 729 to is prime factors you see that 729 = 3^6. So either:

729 = 3^6 (a = 3 and n= 6), or
729 = 9^3 (a = 9, n = 3), or
729 = 27^2 (a = 27, n = 2)

The second coefficient is n*a^(n-1)*b = 2916, so let's see if any of the combinations of a and n work:

For a = 3, n=6:
6*3^5*b = 2916, or b = 2

For a = 9, n = 3:
3*9^2*b = 2916, or b = 12

For a = 27, n = 2:
2*27*b = 2916, or b = 54

Now try these combinations of a, b, and n into the equation for the third coefficient, which you know equals 4860, and see if they check out. The equation of the third coefficient is:

n*(n-1)/2 * a^(n-2) * b^2 = 4860.

For a = 3, b =2, n = 6:
6*5/2 * 3^4 *2^2 = 4860, so this checks out.

For a = 9, b = 12, n=3:
3*2/2 *9*12^2 = 3888, so this is not correct.

For a = 27, b = 54, n = 2:
2*1/2 * 27^0 *54^2 = 2916, so this is also not correct.

Hence the only answer is: a = 3, b =2, and n = 6.

Ebaines, that's what I did to get the answers, thanks anyway both of you!