Binomial Theorem (II) replied by ^tamago^ @ Mon, 16 Jun 2008 17:20:27 +0800

^tamago^ — Mon, 16 Jun 2008 17:20:27 +0800

Binomial Theorem (II) replied by Uncertain @ Mon, 16 Jun 2008 16:57:34 +0800

Uncertain — Mon, 16 Jun 2008 16:57:34 +0800

Originally posted by secretliker:

Huh? What are you talking about? I think you misunderstood me.

This was what you wrote:

The bolded part is wrong. That's all.

Sorry sercetliker. Just misread the lines. Ya it is my mistake.

Binomial Theorem (II) replied by Y_Shun @ Sun, 15 Jun 2008 12:19:43 +0800

Y_Shun — Sun, 15 Jun 2008 12:19:43 +0800

chill man

most impt is TS gets the concept/idea of how to do it

Binomial Theorem (II) replied by secretliker @ Fri, 13 Jun 2008 17:59:16 +0800

secretliker — Fri, 13 Jun 2008 17:59:16 +0800

Huh? What are you talking about? I think you misunderstood me.

This was what you wrote:

(1+x)(a-bx)^12 = (1+x)(a/b-x)^12 (1/b)^12

The bolded part is wrong. That's all.

Binomial Theorem (II) replied by Uncertain @ Thu, 12 Jun 2008 10:58:55 +0800

Uncertain — Thu, 12 Jun 2008 10:58:55 +0800

Originally posted by secretliker:

Solving using your working,

495a^4b^8 - 792a^5b^7 = 0

495a^4b^8 = 792a^5b^7

5/8 * b = a

a/b = 5/8

Uncertain's mistake in line 1 isn't significant though.

(1+x)(a-bx)^12 = (1+x)[b(a/b-x)]^12 = (1+x)(a/b-x)^12 (b)^12

So u think u are damn clever? my mistake? is it a mistake to begin with?

I write more for TS to understand the concept. In fact, this qn only needs 3 lines to solve <-- can only be understood if and only if the reader is a math pro.

Binomial Theorem (II) replied by secretliker @ Thu, 12 Jun 2008 01:07:27 +0800

secretliker — Thu, 12 Jun 2008 01:07:27 +0800

Solving using your working,

495a^4b^8 - 792a^5b^7 = 0

495a^4b^8 = 792a^5b^7

5/8 * b = a

a/b = 5/8

Uncertain's mistake in line 1 isn't significant though.

(1+x)(a-bx)^12 = (1+x)[b(a/b-x)]^12 = (1+x)(a/b-x)^12 (b)^12

Binomial Theorem (II) replied by Uncertain @ Sun, 08 Jun 2008 22:05:07 +0800

Uncertain — Sun, 08 Jun 2008 22:05:07 +0800

(1+x)(a-bx)^12 = (1+x)(a/b-x)^12 (1/b)^12

For (a/b-x)^12,

coefficient of x^7 = 12C7 (a/b)^5 (-1)^7

coefficient of x^8 = 12C8 (a/b)^4 (-1)^8

Therefore,

Coefficient of x^8 = (1/b)^12 [ 12C7(a/b)^5(-1) + 12C8(a/b)^4 ] = 0

a/b = 12C8 / 12C7 = 5/8

Binomial Theorem (II) replied by SBS261P @ Sun, 08 Jun 2008 16:41:20 +0800

SBS261P — Sun, 08 Jun 2008 16:41:20 +0800

729 or 792? check again...

Binomial Theorem (II) replied by bonkysleuth @ Sun, 08 Jun 2008 12:23:50 +0800

bonkysleuth — Sun, 08 Jun 2008 12:23:50 +0800

In the expansion of (1 + x)(a - bx)^12, where ab is not equal to 0, the coefficient of x^8 is zero. Find in its simplest form the value of the ratio a/b.

I expanded (a - bx)^ 12 using binomial theorem. So you get

(+....-729a^5b^7x^7 + 495a^4b^8x^8 ...+)

Now you'll have (1 + x)(+....-729a^5b^7x^7 + 495a^4b^8x^8 ...+)

Whe you expand it out, you'll get 495a^4b^8 - 729a^5b^7 = 0. I don't think i made any careless mistakes in the expansion as I did it numerous times, but somehow or rather, I still can't derive the answer. It seems awkward (the equation). Or is there an alternative to solving this question? By the way, you should get 5/8 in the end.

Recent Posts in 'Binomial Theorem (II)' | sgForums.com