In the expansion of (1+x)(a-bx)12, where ab≠0, the coefficient of x8 is zero.

[audrey.1021]

In the expansion of (1+x)(a-bx)¹², where ab≠0, the coefficient of x⁸ is zero. Find, in its simplest form, the value of ratio a/b.

So I tried comparing coefficients, and I got that I'd need to expand an x⁷ and x⁸ to get the total coefficient of x⁸ in order to equate that to zero. But the fractions just didn't work out.

That method certainly does work. Please show your work. What did you get for the coefficient of x⁷ in (a- bx)¹²? What did you get for the coefficient of x⁸?

I came across the same problem. I am stuck at this step too. The coefficient I got for x⁷ is -792a⁵b⁷ and 495a⁴b^8. I tried to find the ratio but it doesn't seem to work. Can you please explain how to find the ratio of the unknowns?

Thanks

 

[Ishuda]

I came across the same problem. I am stuck at this step too. The coefficient I got for x⁷ is -792a⁵b⁷ and 495a⁴b^8. I tried to find the ratio but it doesn't seem to work. Can you please explain how to find the ratio of the unknowns?

Thanks

Possibly it might be easier to let c=a/b and then look at
(1+x)(c - x)¹²

EDIT: Hint; ₁₂C₆ = (7/6) ₁₂C₅

 

[stapel]

I came across the same problem. I am stuck at this step too. The coefficient I got for x⁷ is -792a⁵b⁷ and 495a⁴b^8. I tried to find the ratio but it doesn't seem to work.

Please reply showing your steps, so we can see where things are bogging down. Thank you!

 

[audrey.1021]

Binomial Theorem

Please reply showing your steps, so we can see where things are bogging down. Thank you!

I expanded (1+x)(a-bx)¹² and I got -792a⁵b⁷x⁷+495a⁴b⁸x⁸. The coefficient of x⁸ equals 0, -792a⁵b⁷x⁷+495a⁴b⁸x⁸=0. I am not sure what to do next.

 

[stapel]

I expanded (1+x)(a-bx)¹² and I got -792a⁵b⁷x⁷+495a⁴b⁸x⁸.

Are you saying that these two terms were your entire expansion? :shock:

 

[audrey.1021]

Are you saying that these two terms were your entire expansion? :shock:

Sorry, I made a mistake in explaining just now. I mean that -792a⁵b⁷+495a⁴b⁸is my coefficient of x⁸ in the expansion of (1+x)(a-bx)¹². I did not multiply out the other terms since we only need to find the coefficient of x⁸ in the expansion of (1+x)(a-bx)¹²​.

 

[Ishuda]

Sorry, I made a mistake in explaining just now. I mean that -792a⁵b⁷+495a⁴b⁸is my coefficient of x⁸ in the expansion of (1+x)(a-bx)¹². I did not multiply out the other terms since we only need to find the coefficient of x⁸ in the expansion of (1+x)(a-bx)¹²​.

495a⁴b⁸ - 792a⁵b⁷ = \(\displaystyle a^5\, *\, b^7\, *\, [ 495\, *\, ?\, -\, 792\, *\, ?]\) = 0

 

[audrey.1021]

495a⁴b⁸ - 792a⁵b⁷ = \(\displaystyle a^5\, *\, b^7\, *\, [ 495\, *\, ?\, -\, 792\, *\, ?]\) = 0

I solved it. I let c= a/b in (1+x)(c - x)¹². -792c⁵+495c⁴ equals to 495c⁴=792c⁵. c=a/b, a/b= 495/792 (5/8).

But how does ₁₂C₆ = (7/6) ₁₂C₅ help in solving the problem? Another way to do it is to do a⁵*b⁷*(495a⁴-792b⁸)=0, is that right?

 

[pka]

I expanded (1+x)(a-bx)12 and I got -792a5b7x7+495a4b8x8. The coefficient of x8 equals 0, -792a5b7x7+495a4b8x8=0. I am not sure what to do next.

See this expansion.

 

[Ishuda]

I solved it. I let c= a/b in (1+x)(c - x)¹². -792c⁵+495c⁴ equals to 495c⁴=792c⁵. c=a/b, a/b= 495/792 (5/8).

But how does ₁₂C₆ = (7/6) ₁₂C₅ help in solving the problem? Another way to do it is to do a⁵*b⁷*(495a⁴-792b⁸)=0, is that right?

I could claim it was just an example but the truth is I can't subtract correctly. The coefficient for x⁸ is the 8th term in the expansion [for (c-x)¹²] which is c⁴₁₂C₈. The coefficient for x⁷ is -c⁵₁₂C₇. That gives
c⁴ ₁₂C₈ - c⁵ ₁₂C₇.
as the coefficient for x⁸ in expansion for the given expression.

So my hint should have been
₁₂C_{8 = (5/8) 12}C₇

Another way to do it is
Coeff for x8 = -792 a⁵ b⁷ + 495 a⁴ b⁸ = a⁴ b⁸ [ -792 (a/b) + 495 ] = 0
or
(a/b) = 495/792
Oh stick! There's another mistake in one of my hints. Off to the corner goes me.

 

[pka]

I expanded (1+x)(a-bx)¹² and I got -792a⁵b⁷x⁷+495a⁴b⁸x⁸. The coefficient of x⁸ equals 0, -792a⁵b⁷x⁷+495a⁴b⁸x⁸=0. I am not sure what to do next.

\(\displaystyle \left( {a - bx} \right)^{12}= \left( \cdots \binom{12}{7}(a)^5(-bx)^7+\binom{12}{8}(a)^4(-bx)^8 \cdots \right)\)

\(\displaystyle (x+1)\left( \cdots -792a^5b^7x^7+495a^4b^8x^8 \cdots \right)\)
\(\displaystyle \left( \cdots -792a^5b^7x^8+495a^4b^8x^8 \cdots \right)\) SEE HERE