Prove using the principle of mathematical induction that:
1.3.5.7....(2n-1).n! = (2n)!/2^n
(n+1)(n+2)....(2n-1)2n = (2n)!/n!
Thankyou so much!
Hi Christina,
These are a little subtle...
$\displaystyle \color{red}\ 1\$
$\displaystyle \color{blue}[1(3)5(7)9........(2n-1)]n!=\frac{(2n)!}{2^n}$ ?
If this formula is valid, then the following will be true
$\displaystyle \left(1(3)5....(2k-1)[2(k+1)-1]\right)(k+1)!=\frac{[2(k+1)]!}{2^{k+1}}$
The original formula is utilised to attempt this proof, because if
being true for n causes the statement (the formulated hypothesis)
to be true for n+1 (the next term), then this means that if the hypothesis is valid for the first term, then this causes the hypothesis to be true for the 2nd, causing it to be true for the 3rd, causing it to be true for the 4th, causing......an endless chain of cause and effect.
That is what the inductive step attempts to do,
to discover if an endless chain of cause and effect exists.
We do this by trying to prove the following in this case...
For the next k (or n), the formula gives, in terms of the current k (or n)
$\displaystyle [1(3)5...(2k-1)(2[k+1]-1)](k+1)!=\frac{[2(k+1)]!}{2^{k+1}}$ ?
$\displaystyle \frac{(2k)!}{2^k}(2[k+1]-1)(k+1)=\frac{[2(k+1)]!}{2^{k+1}}$ ? using (k+1)!=(k+1)k!
$\displaystyle \frac{(2k+1)!(k+1)}{2^k}=\frac{2(2k+1)!(k+1)}{2^{k +1}}=\frac{(2k+2)(2k+1)!}{2^{k+1}}=\frac{(2k+2)!}{ 2^{k+1}}$
Therefore if the hypothesis is true for some n, it will definately be
true for the next value of n.
Hence if it is true for the first value of n of interest, it will be true for all
following natural numbers.
2
$\displaystyle \color{blue}\ (n+1)(n+2)....(2n-1)(2n)=\frac{(2n)!}{n!}$ ?
To apply induction to this (the equation is obvious if it is considered in terms of factorials in general!!),
first consider how this looks...
n=1...... (1+1)=2(1)
n=2.......(2+1)(2+2)=3(4)=4!/2!
n=3.......(3+1)(3+2)(3+3)==4(5)6=6!/3!
Hence, there is an extra factor as n increases by 1.
Therefore, we will be trying to prove if
$\displaystyle ([k+1]+1)([k+1]+2)...([2k+1]-1)([2k+1]-1+1)=\frac{(2[k+1])!}{(k+1)!}$ ?
$\displaystyle (k+2)(k+3)...(2k-1)(2k)(2k+1)(2k+2)=\frac{(2k+2)!}{(k+1)!}$ ?
$\displaystyle \frac{(2k)!}{k!}\ \frac{(2k+2)(2k+1)}{(k+1)}= \frac{(2k+2)!}{(k+1)!}$
The inductive process is proven, just test it for an initial value.