Bionomial Expansions help :)

[n3rdwannab3]

1. The problem statement, all variables and given/known data

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.




2. Relevant equations



3. The attempt at a solution

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.


Any help is appreciated :)

[Ray Vickson]

1. The problem statement, all variables and given/known data

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.




2. Relevant equations



3. The attempt at a solution

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.


Any help is appreciated :)

You have formulas for C(n,r) and C(n,r+1), so you can compute and simplify the ratio C(n,r)/C(n,r+1), and set that equal to 6/14. That is one equation for n and r. Do the same for n and (r+1). Be careful to simplify the ratio as much as you can before proceeding!

RGV

[SammyS]

Let's wait and hope that a Moderator splits this up, before we get too far into helping.

[berkeman]

It is several questions at once (please don't do that), but maybe we can handle them at once since the OP did post work on each?

[SammyS]

It is several questions at once (please don't do that), but maybe we can handle them at once since the OP did post work on each?
Okee-Dokee.

[berkeman]

Okee-Dokee.

Sorry Sammy. You are a huge helper. Should this be split up?

[SammyS]

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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.
...

Any help is appreciated :)

nCr is given by:

\displaystyle _{n}C_{r}=\frac{n!}{r\,!\,(n-r)!}

So, for instance: \displaystyle \frac{_{n}C_{r-1}}{_{n}C_{r}}=\frac{\displaystyle \frac{n!}{(r-1)!\,(n-r+1)!}}{\displaystyle \frac{n!}{r\,!\,(n-r)!}}=\frac{r\,!\,(n-r)!}{(r-1)!\,(n-r+1)!}=\frac{r}{n-r+1}=\frac{6}{14}

Similarly: \displaystyle \frac{_{n}C_{r}}{_{n}C_{r+1}}=\frac{14}{21}

It's pretty easy to solve these for n and r.

[SammyS]

Sorry Sammy. You are a huge helper. Should this be split up?
No, It'll be OK.

[berkeman]

No, It'll be OK.

:smile: