Induction Q (1 Viewer)

[Sparcod]

It's a 1999 HSC question.
Prove
(n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]

Thanks to anyone who helps.

 

[pLuvia]

Oh I've seen this one before, don't think from hsc question, but probably something similar I'll try it

 

[Sparcod]

Oh I've seen this one before, don't think from hsc question, but probably something similar I'll try it

It looks easy but i cant get it.

 

[gman03]

It's a 1999 HSC question.
Prove
(n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]

Thanks to anyone who helps.

Let P(n) == (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]

P(1): LHS = 2, RHS = 2 * 1 = 2. so P(1) true.

Assume P(n) is true. i.e. (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]

For P(n+1), RHS = 2^(n+1) [1x3x...x(2n-1)x(2n+1)]

LHS = (n+2)(n+3)...2n(2n+1)(2n+2)
= (n+1)(n+2)...(2n-1)2n * (2n+1)(2n+2) / (n + 1)
= (n+1)(n+2)...(2n-1)2n * 2(2n+1)

by induction,

= 2^n [1x3x...x(2n-1)] * 2(2n+1)
= 2^(n+1) [1x3x...x(2n-1)x(2n+1)]
= RHS

Therefore P(n+1) true, hence true for n >= 1

 

[pLuvia]

S_n = (n+1)(n+2)...(2n-1)2n= 2ⁿ [1x3x...x(2n-1)]

Show n = 1 is true
1=1
.: S₁ is true
Assume n=k
S_k = (k+1)(k+2)...(2k-1)2k= 2^k [1x3x...x(2k-1)]

Let n = k+1
S_k+1 = (k+1)(k+2)...(2k-1)2k . (2k+1)(2k+2) = 2^k [1x3x...x(2k-1)] . (2k+1)(2k+2)

= 2^k [1x3x...x(2k-1)] . (2k+1)(2k+2)
= 2^k+1 [1x3x...x(2k-1)] . (2k+1)(k+1)
=

This is the furtherest my knowledge takes me

 

[Sparcod]

Use P?? Sounds weird but creative.
Thanks.

 

[gman03]

Let n = k+1
S_k+1 = (k+1)(k+2)...(2k-1)2k . (2k+1)(2k+2) = 2^k [1x3x...x(2k-1)] . (2k+1)(2k+2)

Becareful there

 

[pLuvia]

I don't see where I went wrong

 

[gman03]

S_n = (n+1)(n+2)...(2n-1)2n

so S_k+1 = ((k+1)+1)((k+1)+2)...(2(k+1)-1)2(k+1)
= (k+2)(k+3)...(2k+1)(2k+2)

essentially, you included an extra (k+1) in your working

 

[Riviet]

gman is correct.

I used to do that mistake too.

 

[Sparcod]

Thanks guys.
Be careful with those 'k+1's

 

[DraconisV]

Thanks guys.
Be careful with those 'k+1's

True i tend to do that alot.

Like with this one

[2k-1] then k+1 = [2k] instead of [2(k+1)-1] = [2k+1]

tricky stuff, but after youve done it all you get used to it, and the other two types of induction are sinch as compared to the first type(sums).

 

[Riviet]

True i tend to do that alot.

Like with this one

[2k-1] then k+1 = [2k] instead of [2(k+1)-1] = [2k+1]

tricky stuff, but after youve done it all you get used to it, and the other two types of induction are sinch as compared to the first type(sums).

Is it me or did you get it totally wrong? I thought it was the other way around, 2k is wrong and 2k+1 is right.

e.g 1: the k+1th term in (k-1)(k)(k+1) would be k(k+1)(k+2)]

e.g 2: the k+1th term in (2k-1)(2k+3)(2k+7) would be [2(k+1)-1][2(k+1)+3][2(k+1)+7] = (2k+1)(2k+5)(2k+9)

I think you confused yourself.

 

[Sparcod]

Very controversial induction question.