- #1
MarkFL
Gold Member
MHB
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On another forum the following problem was posted:
"Prove with mathematical induction that
$\displaystyle \frac{(n+1)(n+2)...(2n)}{2^n}$
is an integer when $\displaystyle n\in\mathbb{N}$
My solution:
I chose to write the induction hypothesis $\displaystyle P_n$ after looking at the first several statements:
$\displaystyle \frac{(2n)!}{n!}=(2n-1)!2^n$
We easily see that $\displaystyle P_1$ is true, so next I defined:
$\displaystyle \mu(n)=\frac{(2(n+1))!}{(n+1)!}-\frac{(2n)!}{n!}=\frac{(2(n+1))!-(n+1)(2n)!}{(n+1)!}=\frac{(2n)!((2n+2)(2n+1)-(n+1))}{(n+1)!}=$
$\displaystyle \frac{(2n)!}{n!}(2(2n+1)-1)=\frac{(2n)!}{n!}(4n+1)=(2n-1)!2^n(4n+1)$
Now, adding $\displaystyle \mu(n)$ to both sides of $\displaystyle P_n$ there results:
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^n+(2n-1)!2^n(4n+1)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^n(4n+2)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^{n+1}(2(n+1)-1)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2(n+1)-1)!2^{n+1}$
We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$ thereby completing the proof by induction.
"Prove with mathematical induction that
$\displaystyle \frac{(n+1)(n+2)...(2n)}{2^n}$
is an integer when $\displaystyle n\in\mathbb{N}$
My solution:
I chose to write the induction hypothesis $\displaystyle P_n$ after looking at the first several statements:
$\displaystyle \frac{(2n)!}{n!}=(2n-1)!2^n$
We easily see that $\displaystyle P_1$ is true, so next I defined:
$\displaystyle \mu(n)=\frac{(2(n+1))!}{(n+1)!}-\frac{(2n)!}{n!}=\frac{(2(n+1))!-(n+1)(2n)!}{(n+1)!}=\frac{(2n)!((2n+2)(2n+1)-(n+1))}{(n+1)!}=$
$\displaystyle \frac{(2n)!}{n!}(2(2n+1)-1)=\frac{(2n)!}{n!}(4n+1)=(2n-1)!2^n(4n+1)$
Now, adding $\displaystyle \mu(n)$ to both sides of $\displaystyle P_n$ there results:
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^n+(2n-1)!2^n(4n+1)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^n(4n+2)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!2^{n+1}(2(n+1)-1)$
$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2(n+1)-1)!2^{n+1}$
We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$ thereby completing the proof by induction.