induction question - please help! (1 Viewer)

[frangipani13]

okay well usually im fairly good at induction but im having a lot of trouble with this...due in on tuesday to my teacher. its from the 1999 hsc paper:

Prove by induction that for all integers >= 1:
(n+1)(n+2)...(2n-1)2n = 2^n [1x2x...x(2n-1)]

its the ... on both sides thats confusing me. its true for n=1 but im stuck on proving it for n=k+1

(k+2)(k+3)...(2k+1)(2k+2) = 2^(k+1) [1x2x...x(2k)]

thanks...any help is much appreciated!

 

[pLuvia]

Let n=1
S₁ = 1 = 1
S₁ is true

Assume n=k
.: (k+1)(k+2)...(2k-1)2k = 2^k[1x2x...x(2k-1)]

Let n=k+1
.: (k+1)(k+2)...(2k-1)2k x (2k+2)(2k+1) = 2^k[1x2x...x(2k-1)] x (2k+2)(2k+1)
= 2^k[1x2x...x(2k-1)] x (2k+2)(2k+1)
=

Well this is all I can get up to If I can do more I'll post it up

 

[richz]

hi frangipani13,

ok

Assume n=k
.: (k+1)(k+2)...(2k-1)2k = 2^k[1x2x...x(2k-1)]

Prove true for n=k+1
(k+2)(k+3)....(2k-2)(2k-1)2k.2k(2k+1)(2k+2)= 2^k+1[1x2x...x(2k+1)]

the term in bold can be simplified, 2k+2=2(k+1)

LHS=2(k+1)(k+2)(k+3)....(2k-1)2k.2k(2k+1)

= 2.2^k[1x2x...x(2k-1)]2k(2k+1)

ok now, divide both sides by 2k+1

u see now

 

[klaw]

The question's wrong.
When n=3,
LHS=4*5*6=120
RHS=8*5!=960

Confirmed: http://www.boardofstudies.nsw.edu.a...hs/99MATH34.PDF

 

[MAICHI]

I don't know, I got to an expression which is 2n+1 short, there is something wrong with the question. If you do some algebra you can get this expression from the original question: 2n = (2^n)*n!, which clearly is not possible except for n=1.

 

[MAICHI]

Confirmed: http://www.boardofstudies.nsw.edu.a...hs/99MATH34.PDF

I can't access that for some reason. But so the actual HSC question was wrong?! Freaky.

 

[klaw]

No, her teacher copied the HSC Q wrong

 

[Riviet]

Why did you guys multiply by 2 terms instead of 1
i.e (2k+1)(2k+2)
shouldn't it just be multiplied by (2k+2) since you multiply by the k'th term i.e:
2(k+1)=2k+2
If i am right then check this out:
Assume true for n=k
i.e (k+1)(k+2)(k+3)...(2k-1)2k=2^k[1x2x3x...x(2k-1)]
Prove for n=k+1
i.e (k+2)(k+3)(k+4)...(2k-1)2k(2k+2)=2^k+1[1x2x3x...x(2k+1)]
LHS=(k+2)(k+3)(k+4)...(2k-1)2k(2k+2)
=2[(k+1)(k+2)(k+3)...(2k-1)2k]
=2^k[1x2x3x...x(2k-1)][2(2k+1)] by the assumption
=2^k+1[1x2x3x...x(2k-1)](2k+1)
=RHS!

 

[frangipani13]

no my teacher photocopied a page from the success one book on induction q's...the marker's comments i read said the vast majority of hsc students couldn't do it properly because it involved production rather than addiction. (unfortunately didnt have the solution) im sure the question's right.

 

[richz]

well, call the maths book to fuck a duck, check out the website, it even says 3.

 

[klaw]

Why did you guys multiply by 2 terms instead of 1
i.e (2k+1)(2k+2)
shouldn't it just be multiplied by (2k+2) since you multiply by the k'th term i.e:
2(k+1)=2k+2
If i am right then check this out:
Assume true for n=k
i.e (k+1)(k+2)(k+3)...(2k-1)2k=2^k[1x2x3x...x(2k-1)]
Prove for n=k+1
i.e (k+2)(k+3)(k+4)...(2k-1)2k(2k+2)=2^k+1[1x2x3x...x(2k+1)]
LHS=(k+2)(k+3)(k+4)...(2k-1)2k(2k+2)
=2[(k+1)(k+2)(k+3)...(2k-1)2k]
=2^k[1x2x3x...x(2k-1)][2(2k+1)] by the assumption
=2^k+1[1x2x3x...x(2k-1)](2k+1)
=RHS!

No. 2n+1 is an AP, where the common difference is 2. Therefore the number increases by 2 every time. Therefore 1*2*3*... does not fit in the series. You can see it's wrong just from the Q.

 

[Riviet]

Oo i see, yeah you're right it goes up by one at the start and ends up going by 2 lol. :uhhuh:

 

[richz]

check my working, i finished it i think