Binomial theorem, with 2 unknowns

• Feb 3rd 2006, 01:02 AM
freswood
Binomial theorem, with 2 unknowns
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!
• Feb 3rd 2006, 02:33 AM
ticbol
Quote:

Originally Posted by 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!

Binomial Theorem,
(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.

----------------------------------------------------------
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?
• Feb 3rd 2006, 05:17 AM
phgao
Wow, this is what I will be starting in maths very soon! Can you explain why for me ticbol? If it's too compilcated then forget it but if not :D ...

Thanks!
• Feb 3rd 2006, 09:43 AM
ThePerfectHacker
Notice that $n!=n(n-1)(n-2)...(2)(1)$ can be written as $n!=n[(n-1)(n-2)...(2)(1)]$ thus,
$n!=n(n-1)!$
$\frac{n!}{(n-1)!}$ can be written as by previous demonstration as,
$\frac{n(n-1)!}{(n-1)!}$ cancel, thus,
$n$.

By similar reasoning you can express,
$\frac{n!}{(2!)(n-2)!}$ as
$\frac{n(n-1)(n-2)!}{(2)(n-2)!}$
cancel, thus,
$\frac{n(n-1)}{2}$
Q.E.D.
• Feb 3rd 2006, 01:52 PM
freswood
Thanks so much for the explanation!
• Feb 3rd 2006, 03:03 PM
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
• Feb 3rd 2006, 05:05 PM
ticbol
Quote:

Originally Posted by 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

Yes, you're on track, but from there, it's guessing time because the givens are not enough.

Your "coefficient of the 2nd terms" means the (nC1)*(whatever is numerical in the [(2a)^(n-1)]*[-1]).
Your "coefficent of the 2nd term" is not the binomial coefficient only---which is nC1 here, which is "n" only.

(2a -1)^n
Second term is [n]*[(2a)^(n-1)]*[-1].
Simplifying that,
= (-n)[2^(n-1)]*[a^(n-1)]
= (-n)[(2^n)/2][a^(n-1)]
= [-n(2^n)/2][a^(n-1)]
And the "coefficient" here is the one without the variable "a".
The "coefficient" here is [-n(2^n)/2].

Equate that to (-192),
-n(2^n)/2 = -192
n(2^n) = 384 -------------(1)

That is the same as your 2^(n-1) = 192/n.

We make a guess, or we make trial and error.

If n = 4,
n(2^n) = 384
4(2^4) =? 384
4(16) =? 384
64 =? 384
No.

If n = 5,
n(2^n) = 384
5(2^5) =? 384
5(32) =? 384
160 =? 384
No.

If n = 6,
n(2^n) = 384
6(2^6) =? 384
6(64) =? 384
384 =? 384

2^(n-1) = 192/n
2^(6-1) =? 192/6
2^5 =? 32
32 =? 32
Yes, so, OK.

And the 2nd term is
n[(2a)^(n-1)][-1]
= 6[(2a)^(6-1)][-1]
= -6[(2a)^5]
= -6[32a^5]
= -192a^5

-------------------------------------------
Maybe next time, if you have another question separate from that/those posted above, you post or ask it in another posting.
Don't be shy, you can post as many separate postings as you want. Make one posting only per one question/problem if possible.
• Feb 3rd 2006, 05:23 PM
freswood
Isn't there a way other than trial and error?
• Feb 3rd 2006, 06:36 PM
ticbol
Quote:

Originally Posted by freswood
Isn't there a way other than trial and error?

How about looking? Just by looking. :-)
That may be bragging, but it works for some guys.
Looking and fast thinking or mental exercise.

So a guy is stopped by n(2^n) = 384.
There might be ways to compute for n by logarithms, "e", limits [ :-) ], or other Math stuff, but he couldn't get it at the moment.
Heck, why not do it the "computer way" ---------trial and error, or comparison, or iteration, or ......

He thought of 384 in powers of 2, since there is a 2^n on the lefthand side.
2^4 = 16, so 4*16 = 64
2^5 = 32, so 5*32 = 160
2^6 = 64, so 6*64 = 384
Heck, that is it. Why not write it down? Gotta show some computations.

-------------------
No, I don't know of any other way yet.
• Feb 4th 2006, 12:30 AM
CaptainBlack
Quote:

Originally Posted by ticbol
How about looking? Just by looking. :-)

So a guy is stopped by n(2^n) = 384.
There might be ways to compute for n by logarithms, "e", limits [ :-) ], or other Math stuff, ....

Lambert's W function can be used to solve this, but I would not recommend
it, as the W-function is only considered elementary by its enthusiasts.

But as we know that n must be an integer, and that the powers of 2 are
2, 4, 8, 16, 32, 64, 128, 256, .., it is obvious that 2^6=64 is what we want here.

(some of us -because of our professions- have the powers of 2 engraved
on our memories in characters 20m tall :eek: )

RonL
• Feb 4th 2006, 03:17 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Lambert's W function can be used to solve this, but I would not recommend
it, as the W-function is only considered elementary by its enthusiasts.

But as we know that n must be an integer, and that the powers of 2 are
2, 4, 8, 16, 32, 64, 128, 256, .., it is obvious that 2^6=64 is what we want here.

(some of us -because of our professions- have the powers of 2 engraved
on our memories in characters 20m tall :eek: )

RonL

$n 2^n = 384$ may be rearranged:

$n \ln(2) e^{n \ln(2)}= \ln(2) \times 384$,

Which from the definition of Lambert's $W$ has solution:

$n=\frac{W(384 \ln(2))}{\ln(2)}=6$

RonL