Two Binomial Questions (1 Viewer)

[GaDaMIt]

Given that (1 + x + ax²)⁸ = 1 + 8x + 4x² + bx³ + ....
Find a and b

AND

3 consecutive coefficients in the expansion (1 + x)ⁿ are in the ratio 6 : 3 :1
Find the value of n and state which terms possess these coefficients



I can't do either of those

And another question where im just asking for the answer being checked..

Find the ratio of (r + 1)th term to the rth term in the expansion of (a + b)ⁿ

I got b(n - r) : a(r + 1)

I did this by assuming the identity T(r+1) = ⁿC_r+1 (a)^n-r-1 (b)^{r + 1} instead of ⁿC_r etc.. as those are the two different identities you can apply...

 

[Affinity]

First question:
I'll present 2 ways of thinking which amounts to the same idea:
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use the binomial formula 2 times:

[(1+x) + ax^2]^8 = SUM_{r=0 to 8} [8 Choose r] * (ax^2)^r * (1+x)^{8-r}

Notice that for terms where r >= 2, all powers of x will atleast be 4, so one needs only to consider the terms where r = 0,1 That is

[ (1+x)^8 ] + [ 8ax^2 * (1+x)^7 ]
= [1 + 8x + 28x^2 + 56x^3 + ... +x^8] + [8ax^2 + 56ax^3 + ...]
= 1 + 8x + (28 + 8a)x^2 + (56 + 56a)x^3 + ...
so a=-3 and b = -112
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2nd way, one thinks: So how does one get x^2 terms from 1,x and ax^2
You can have:
1*1*1*1*1*1*1*ax^2
1*1*1*1*1*1*ax^2*1
...
ax^2 * 1 * 1 * 1 ...

And

1*1*1*1*1*1*x*x
1*1*1*1*1*x*1*x
....

and you get the same results.
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Question 2.
One knows that
[n choose r]/[n choose (r+1)]= [ n! / (r!(n-r)!) ] / [n!/((r+1)!(n-r-1)!)]
= (r+1)/(n-r)

The question whats n and k such that

[n choose k] : [n choose k+1] : [n choose k+2] = 6:3:1

therefore
[n choose k] / [n choose k+1] = 2
[n choose k+1] / [n choose k+2] = 3

(k+1)/(n-k) = 2
(k+2)/(n-k-1) = 3

k+1 = 2n - 2k
k+2 = 3n -3k -3

solving gives n = 11, k = 7
so x^7, x^8 and x^9 are the relevant terms

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Question 3:
what you did is correct for your interpretation of T(r) (or equivalently T(r+1)). But it's more usual to have T(r) = nCr * a^r * b^(n-r) instead of what you had, which is T(r) = nCr *a^(n-r) *b^r

 

[GaDaMIt]

Last set of binomial theorum questions...
( i) using n C r notation write down the expansions of (1 + x)^5 and (1 - x)^5

(ii) simplify (1 + x)^5 and (1 - x)^5

(iii) by making an appropriate subsitiution for x, show 5C1 + 5C3 + 5C5 = 2^4

(iv) similarly prove, nC1 + nC3 + nC5 + nC7 + ........ = 2^(n-1)
i only need help with this last one, the others i just provided for reference...



AND


( i ) use n C r notation to expand (1+x)^2n

(ii) by making a suitable substitution for x, prove 2nC0 + 2nC1 + 2nC2 + ....... + 2nC2n = 4^n


EDIT: also for the first set of questions, question iii).. im not sure i did it how they expected, i mean sure i wrote it up in that form but then i just did it as 5C1 + 5C3 + 5C5 = 5 + 10 + 1 = 16 = 2^4; im thinking they expected something else cause i dont see how that interlinks with the last question, as implied by "similarly"..

 

[SoulSearcher]

For iv), just have n instead of 5, then show like in part iii) that the sum of ⁿC₁ + ⁿC₃ + ⁿC₅ + ... = 2^n-1

To do part iii) like they wanted you to, you would have (1+x)⁵ - (1-x)⁵ = ⁵C₀ + ⁵C₁x + ⁵C₂x² + ⁵C₃x³ + ⁵C₄x⁴ + ⁵C₅x⁵ - (⁵C₀ - ⁵C₁x + ⁵C₂x² - ⁵C₃x³ + ⁵C₄x⁴ - ⁵C₅x⁵), then substitue in x = 1 and simplify it from there until you get the final result. Therefore you can sub in n in place of 5 and solve similarly.

The second question, expand as normal, except in place of n you have 2n, and then sub in x = 1 to prove the result.