Mathematical Induction T_T (1 Viewer)

[scardizzle]

I found this question in one of my school's past papers and I've got absolutely no idea:

Prove By Mathematical Induction:

(n+1)(n+2) ... (2n-1)2n = 2^n x (1 x 3 x ... x (2n-1)) for n >= 9

any ideas?

 

[lolokay]

just replace all the n's with (n+1)../ or k with (k+1) whatever

 

[Drongoski]

Let me try:

 

[scardizzle]

thanks for the reply but i'm still having trouble understanding
for this step where does the 2k come from?
(k+1+1)(k+1+2)...2k(2(k+1)-1)(2k+1))

 

[Drongoski]

thanks for the reply but i'm still having trouble understanding
for this step where does the 2k come from?
(k+1+1)(k+1+2)...2k(2(k+1)-1)(2k+1))

For n=k+1, LHS becomes ([k+1] + 1)([k+1] +2) ... 2k(2[k+1]-1)(2[k+1])

We now wish to show formula holds for n = k+1. But what does this mean? It means the RHS must hold true for n=k+1, i.e. it must look like this:

2^k+1 (1x3x5x . . . 2k x (2[k+1]-1)(2[k+1])

i.e. to say the RHS assumes the form of the original formula with n now replced by 'k+1'

Many beginning students of Math Induction have difficulty in seeing what 'true for n=k+1' mean. To be good at Math Induction, you also need to be good at algebraic fiddling !

Hope this helps

 

[Affinity]

I found this question in one of my school's past papers and I've got absolutely no idea:

Prove By Mathematical Induction:

(n+1)(n+2) ... (2n-1)2n = 2^n x (1 x 3 x ... x (2n-1)) for n >= 9

any ideas?