两道求证题10.12a

Question

两道求证题10.12a

Answer 1

1、(2n)!/(n!2^n)=1*3*5*...(2n-1)
(2n)!=2n(2n-1)(2n-2)(2n-3)(2n-4)...(2n-n+1)(2n-n)(2n-n-1)...2*1
=[2n](2n-1)[2n-2](2n-3)[2n-4]...(2n-n+1)(2n-n)(2n-n-1)...[2]*1
=[2n][2n-2][2n-4]...[2]*(2n-1)(2n-3)(2n-5)...1
∴(2n)!/(n!*2^2)
={[n][n-1][n-2]...[1]*(2n-1)(2n-3)(2n-5)...1}/n!
={[n!]*(2n-1)(2n-3)(2n-5)...1}/n!
=(2n-1)(2n-3)(2n-5)...1
2、1*1!+2*2!+3*3!+...+n*n!=(n+1)!-1
∵(k+1)!=(k+1)*k!=k*k!+k!
∴2!=1*1!+1!
3!=2*2!+2!
4!=3*3!+3!
...
(n+1)!=n*n!+n!
等式两边分别求和,约简,然后将右边多余的1!移到左边即可得出结论.