Bionomial Expansions help :)

n3rdwannab3 · **Joined:** Mon, 6 Feb 2012 02:21:29 UTC **Posts:** 1

I am really stuck on this and I've spent hours trying to figure out these question sets.

A) Three consecutive coefficients in the expansion of (1+x)^n are in the ratio 6:14:21. Find the value of n.

B) Find the independent term in the [2x + 1 - 1/(2x^2)]^6 (independent term is x^0)

C) In the expansion of (1 + ax)^n the first term is 1, the second term is 24x, and the third term is 252x^2. Find the values of a and n

D) In the expansion of (x + a)^3(x - b)^6, the coefficient of x^7 is -9 and there is no x^8 term. Find a and b.

So I have worked on all four of them:
A) I realized that the coefficients are n C r, n C (r+1), and n C (r+2), and that they are in ratio of 6:14:21. However I am not sure how to find n after this step.

B) I tried substituting a variable y { let y = 2x - 1/(2x^2) } to form (y + 1)^6. I could expand it all out and test each term using the general term of binomial expansion, but that would be very tedious. I am wondering if there is a better solution

C) I simplified C) to (a)(n C 1) = 24 and (a^2)(n C 2) = 252. Now I'm stuck on what to do next

D) I've simplified it to 5b^2 -6ab + a^2 = -3 and -2b^2 +a = 0. I'm not sure how to proceed from this.

Again, any help is much appreciated.

Shadow · **Posted:** Mon, 6 Feb 2012 04:24:36 UTC

n3rdwannab3 wrote:

I am really stuck on this and I've spent hours trying to figure out these question sets.

A) Three consecutive coefficients in the expansion of (1+x)^n are in the ratio 6:14:21. Find the value of n.

B) Find the independent term in the [2x + 1 - 1/(2x^2)]^6 (independent term is x^0)

C) In the expansion of (1 + ax)^n the first term is 1, the second term is 24x, and the third term is 252x^2. Find the values of a and n

D) In the expansion of (x + a)^3(x - b)^6, the coefficient of x^7 is -9 and there is no x^8 term. Find a and b.

So I have worked on all four of them:
A) I realized that the coefficients are n C r, n C (r+1), and n C (r+2), and that they are in ratio of 6:14:21. However I am not sure how to find n after this step.

B) I tried substituting a variable y { let y = 2x - 1/(2x^2) } to form (y + 1)^6. I could expand it all out and test each term using the general term of binomial expansion, but that would be very tedious. I am wondering if there is a better solution

C) I simplified C) to (a)(n C 1) = 24 and (a^2)(n C 2) = 252. Now I'm stuck on what to do next

D) I've simplified it to 5b^2 -6ab + a^2 = -3 and -2b^2 +a = 0. I'm not sure how to proceed from this.

Again, any help is much appreciated.

You know that ${n!\over (n-k)!k!}/{n!\over (k+1)!(n-k-1)!}=6/14$ , write this out, cancel some terms, then write out the other ratio information and go from there.

Soroban · **Posted:** Mon, 6 Feb 2012 14:14:54 UTC

Hello, n3rdwannab3!

Quote:

(C) In the expansion of (1 + ax)^n

, the first term is 1,
the second term is 24x

, and the third term is 252x^2.

Find the values of

and

I simplified to: . $(a)(_nC_1) = 24\,\text{ and }\,(a^2)(_nC_2) \:=\: 252.$
Now I'm stuck on what to do next.

Hmmm, the next few steps should be obvious . . .

$(a)(_nC_1) \:=\:24 \quad\Rightarrow\quad an \:=\:24 \quad\Rightarrow\quad a \:=\:\dfrac{24}{n}$ .[1]

$(a^2)(_nC_2) \:=\:252 \quad\Rightarrow\quad (a^2)\left(\dfrac{n(n-1)}{2}\right) \:=\:252$ .[2]

Substitute [1] into [2] and solve for

. . . then solve for

Soroban · **Posted:** Mon, 6 Feb 2012 14:53:41 UTC

Hello again, n3rdwannab3!

Quote:

(B) Find the constant term in . $\left[2x + 1 - \dfrac{1}{2x^2}\right]^6$

I tried substituting a variable

Let $y \:=\: 2x - \frac{1}{2x^2}$ .to form (y + 1)^6.

I could expand it all out and test each term using the general term of binomial expansion,
. . but that would be very tedious. .I am wondering if there is a better solution.

My method is somewhat less tedious . . . but still a long procedure.

$$\text{We have: }\:\left[(2x+1) - \frac{1}{2x^2}\right]^6 \;=\;\underbrace{(2x+1)^6}_{(a)} -\;6\underbrace{\frac{(2x+1)^5}{2x^2}}_{(b)} +\;15\underbrace{\frac{(2x+1)^4}{(2x^2)^2}}_{(c)} -\;20\underbrace{\frac{(2x+1)^3}{(2x^2)^3}}_{(d)} + \hdots$

In (a), the constant term is

In (b), we have: . $\dfrac{(2x)^5 + 5(2x)^4 + 10(2x)^3 + 10(2x)^2 + 5 + 1}{2x^2}$
The constant term is: . $\dfrac{40x^2}{2x^2} \,=\,20$

In (c), we have: .[/c
olor] $\dfrac{(2x)^4 + 4(2x)^3 +6(2x)^2 + 4(2x) + 1}{(2x^2)^2}$
The constant term is:[color=beige] . $\dfrac{16x^4}{4x^4} \,=\,4$

In (d), we have: . $\dfrac{(2x)^3 + 3(2x)^3 + 3(2x) + 1}{(2x^2)^3}$
There is no constant term.
The same is true for the subsequent terms.

Therefore, the constant term is: . $1 - 6(20) + 15(4) \;=\;-59$

S.O.S. Mathematics CyberBoard

Bionomial Expansions help :)

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