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[rumour]

What topic would the following question be in?

It was in my BT+Prob. assessment, but the last bit seems to be a MI.

Does anyone know how to do it?

i.)Write down the formula for the coefficient of x^r in the expansion of (1 + x)^n, where r & n are positive integers & 1 < r < n (less than)

ii.)Let s & t be postive consecutive integers, with t= s + 1.
Show the s^2n + 2nt -1 is divisible by t^2.

iii.)Deduce the 5^20 + 119 is divisible by 36.

 

[acmilan]

i) is definitely binomials

 

[Grey Council]

ii) is induction? not sure, use BT plus induction :-\

 

[CrashOveride]

i thought last 2 were both ind.

 

[Estel]

2.. requires (t-1)^2k I believe, which would require binomial? Yes? No?

 

[:: ck ::]

well of course it requires (t-1)^2k ... it says deduce ..

havent done binomial yet at school [ive done it.. but not in much detail] .. so not sure if its an induction q or not

 

[rumour]

Thanks guys, but can anyone actually do it?

 

[Harimau]

Part 1 is easy.

The Second Part i think is a little more challenging.

What we are required to show is that s^(2n) + 2nt - 1 is divisible by t^2

Now s^(2n) - 1 is clearly divisibly by t^2, just expand the binomiall and you can see for yourself.

the problem lies in the term 2nt/t^2, which simplifies to 2n/t.

So what you need to do is just to prove 2n/t is an integer, which i really don't know what to do.

n and t are both integers, but are not correlated with each other. So i don't know what do from there.

Part 3 is just using the formula derived in part 2. Sub n = 10, and t = 6 and it will work out.

 

[rumour]

I think in part 2, the -1 cancels out.

 

[Harimau]

Originally posted by rumour
I think in part 2, the -1 cancels out.

Thats what i said.

 

[rumour]

Originally posted by Harimau
Thats what i said.

*woops*

Sorry I blanked out there!!!!

I'll check with the answers the teacher just gave us!!!!

 

[CM_Tutor]

None of this is induction - it's all binomial.

Originally posted by rumour
ii.)Let s & t be postive consecutive integers, with t= s + 1.
Show the s^2n + 2nt -1 is divisible by t^2.

iii.)Deduce the 5^20 + 119 is divisible by 36.

s²ⁿ = (t - 1)²ⁿ, as t = s + 1
= ²ⁿC₀t²ⁿ + ²ⁿC₁t^2n-1(-1)¹ + ²ⁿC₂t^2n-2(-1)² + ²ⁿC₃t^2n-3(-1)³ + ... + ²ⁿC_2n-1t^2n-(2n-1)(-1)^2n-1 + ²ⁿC_2nt^2n-2n(-1)²ⁿ
= ²ⁿC₀t²ⁿ - ²ⁿC₁t^2n-1 + ²ⁿC₂t^2n-2 - ²ⁿC₃t^2n-3 + ... - ²ⁿC_2n-1t + ²ⁿC_2n
= t²(²ⁿC₀t^2n-2 - ²ⁿC₁t^2n-3 + ²ⁿC₂t^2n-4 - ²ⁿC₃t^2n-3 + ... + ²ⁿC_2n-2) - 2nt + 1, as ²ⁿC_2n-1 = 2n and ²ⁿC_2n = 1
So, s²ⁿ + 2nt - 1 = t²(²ⁿC₀t^2n-2 - ²ⁿC₁t^2n-3 + ²ⁿC₂t^2n-4 - ²ⁿC₃t^2n-3 + ... + ²ⁿC_2n-2), which is divisible by t²

So, s²ⁿ + 2nt - 1 is divisible by t², as required.

Now, put t = 6 and n = 10. It follows that s = 5, and hence 5²⁰ + 2(10)(6) - 1 = 5²⁰ + 119 is divisible by 6² = 36

 

[rumour]

Thanks CM!