Question

Problem of mathematical induction (MI)
Prove by MI that (n+1)(n+2)(n+3)...(2n) = (2^n) . 1 .3 . 5 ... (2n-1)

Answer 1

I would prefer a more readable version.

Answer 2

Let p(n) be the statement (n+1)(n+2)(n+3) ‧‧‧ (2n) = (2^n) ‧1 ‧ 3 ‧ 5 ‧‧‧ (2n-1)\]
When n =1
LS = 1+1 =2
RS = (2^1)(2-1) = 2
So, p(1) is true.
Assume p(k) is true
When n =k+1
.... Then I can't continue :S

Answer 3

Or it should be I don't understand...
Assume p(k) is true
When n =k+1
LS = (k+2)(k+3) ‧‧‧ (2k)(2k+1)(2k+2) <- I don't understand the last 3 terms ...

Answer 4

If p(k) is true, \[(k+1)\cdot(k+2)\cdots(2k) =2^k\cdot 1\cdots(2k-1)\]
p(k+1) is,\[(k+2)\cdot(k+3)\cdots(2k+2)=2^{k+1}\cdot 1\cdots(2k)\]\[\implies (k+2)\cdots(2k)\cdot(2k+1)\cdot(2k+2)\cdot\frac{(k+1)}{(k+1)} = 2\cdot 2^k\cdot1\cdot2\cdots (2k-1)\cdot 2k\]

Answer 5

:| I think I got it now
2k = n-2 = 2(k+1) -2 = 2k
2k+1 = n-1 = 2(k+1) -1 = 2k+1
2k+2 = n = 2(k+1) = 2k+2
Right?!

Answer 6

Typo, it's supposed to be\[\frac{(k+1)}{(k+1)}\cdot(k+2)\cdots(2k+2)=2\cdot2^{k}\cdot1\cdots(2k+1) \]

Using p(k) on LHS.
\[\text{LHS}=\frac{2(2k+1)(k+1)}{k+1}\cdot2^k\cdot1\cdots(2k-1)\]

Answer 7

I might have missed something? Yes, I missed out the term 2k

Answer 8

Rest is fine.

Answer 9

Is it readable (my solution)? Do you understand it?

Answer 10

Your solution is readable. But you haven't answered my stupid question :|

Answer 11

Yes, you did right.

Answer 12

Actually I did it in this way..
LS = (k+2)(k+3) ‧‧‧ (2k)(2k+1)(2k+2)
= [2(k+1)(k+2)(k+3) ‧‧‧ 2k] (2k+1)
= 2 ‧ (2^k) ‧ 1 ‧ 3 ‧ 5 ‧‧‧ (2k-1)(2k+1)
= 2^(k+1) ‧ 1 ‧ 3 ‧ 5 ‧‧‧ (2k-1)([2(k+1)-1]
= RS

Answer 13

Thanks!!!~ :)