Binomial Theorem Questions (1 Viewer)

[UnrealAnchovies]

Hiya,

I have a few BT questions which I'm not able to complete. They're probably pretty straightforward.

1) In the expansion of (1+x)^20 in ascending order of x show that there are two successive terms with the same coefficient, which is the greatest of any of the coefficients.

2) Prove that in the expansion of (1+2x)^9, when x = 1/3, there are two greatest terms.

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3) By equating the coefficient of x^r on each side of the identities:
(1+x)^(n+1) = (1+x)(1+x)^n and (1+x)^(n+2) = (1+x)^2(1+x)^n
show that n+1Cr = nCr + nCr-1 and n+2Cr = nCr + 2(nCr-1) + nCr-2

Thanks in advance.

 

[Bored Of Fail]

I wish i could care enough t do them, but honestly i cant.

Also, I would need to write it on paper and scan, or use latex, because all the typing will be impossible to understand, and im to lazy for that

 

[Bored Of Fail]

plus it will take like 20-30 to work them out and type it up

 

[UnrealAnchovies]

:\

Well if you could just give me a start, that would be great.

 

[Bored Of Fail]

hopefully someone else will

 

[Aindan]

1) is incorrect. In your calculator type 20C9,20C10, 20C11. There is only one greatest coefficient.
2) Expand using binomial, you should get 9 terms, Sub x = 1/3. and you get teh result. It's a lot of calculator work...
3) Expand using binomials for each part. for example: (1+x)^n = 1 + nC1x + nC2 x^2 +...+ nC(r-1)x^r-1 + nCr x^r+...+x^n. Pay close attention to the values which will multiply to get x^r. Then compare the coefficients of x^r on LHS and RHS and you should get the two results. THe second proving question is just an expansion of the first.

 

[Bored Of Fail]

^ lols, there is a much eaisesr way of doing number 2, you dont evaluate all the coefficents , lol thats just stupid

 

[UnrealAnchovies]

1) is incorrect. In your calculator type 20C9,20C10, 20C11. There is only one greater coefficient.
2) Expand using binomial, you should get 9 terms, Sub x = 1/3. and you get teh result. It's a lot of calculator work...
3) Expand using binomials for each part. for example: (1+x)^n = 1 + nC1x + nC2 x^2 +...+ nC(r-1)x^r-1 + nCr x^r+...+x^n. Pay close attention to the values which will multiple to get x^r. Then compare the coefficients of x^r on LHS and RHS and you should get the two results. THe second proving question is just an expansion of the first.

Thanks Aindan. The wording of question 1 threw me off... It just didn't make any sense.

 

[UnrealAnchovies]

1) is incorrect. In your calculator type 20C9,20C10, 20C11. There is only one greater coefficient.
2) Expand using binomial, you should get 9 terms, Sub x = 1/3. and you get teh result. It's a lot of calculator work...
3) Expand using binomials for each part. for example: (1+x)^n = 1 + nC1x + nC2 x^2 +...+ nC(r-1)x^r-1 + nCr x^r+...+x^n. Pay close attention to the values which will multiple to get x^r. Then compare the coefficients of x^r on LHS and RHS and you should get the two results. THe second proving question is just an expansion of the first.

Double post.

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Yah, the second one has a quicker method, but I don't mind the expansion if I've got the time.

 

[Aindan]

I'm quite rusty, haven't touched 3unit in 4-5 months. Did you use the T(r+1)/T(r) = (n-r+1)/r?

 

[Bored Of Fail]

how easy are binomials aye

 

[Aindan]

how easy are binomials aye

i reckon they are the hardest 3unit topic @_@.

 

[UnrealAnchovies]

I'm quite rusty, haven't touched 3unit in 4-5 months. Did you use the T(r+1)/T(r) = (n-r+1)/r?

I just use: T(r+1)/T(r) > 1

 

[Aindan]

But even if you did get R < 5. How would you know taht by subbing in x = 1/3 you would get the greatest numerical answer. Because x^3 and x^4. If x = 1/3. the total answer of x^3 is 3 times greater than x^4 now.

 

[UnrealAnchovies]

But even if you did get R < 5. How would you know taht by subbing in x = 1/3 you would get the greatest numerical answer. Because x^3 and x^4. If x = 1/3. the total answer of x^3 is 3 times greater than x^4 now.

Subs. r = 3 and r = 4 into T(r+1) and you get the same solution. But it's got me thinking as well... Something to ask my teacher tomorrow I guess.

 

[UnrealAnchovies]

Is T(r+1)/ T(r) = (n-1+r)/r a formula?

 

[hscishard]

But even if you did get R < 5. How would you know taht by subbing in x = 1/3 you would get the greatest numerical answer. Because x^3 and x^4. If x = 1/3. the total answer of x^3 is 3 times greater than x^4 now.

The coeffiecients are different. It could be 8x^3 and 2x^4

 

[hscishard]

Is T(r+1)/ T(r) = (n-1+r)/r a formula?

Yea..
I personally do it T(r)>=T(r-1). Different but same result

 

[UnrealAnchovies]

Yea..
I personally do it T(r)>=T(r-1). Different but same result

>=? I was taught it was just >...

 

[hscishard]

It doesn't really matter. All you're trying to do if to find the greatest term. It would be obvious that if r is a positive integer, there are two greatest terms/coeffiecients. That is why I prefer the =