n的双阶乘除以n+1的双阶乘

湛尚府925证明:2N的阶乘除以2的N次成以N的阶乘=1.3.5.7.......(2n - 1) -
冉昭莘19539992020 ______ 答: (2n)! =1*2*3*...*(2n-1)*2n =1*3*5*...*(2n-1)*2*4*6*...*2n =1*3*5*...*(2n-1)*2^n*(1*2*3*..*n) =1*3*5*...*(2n-1)*2^n*n! 所以(2n)!/(2^n*n!) =1*3*5*...*(2n-1)*2^n*n!/(2^n*n!) =1*3*5*...*(2n-1)

湛尚府925如何证明(2n)的阶乘/2的n次方乘以n的阶乘=1.3.5.……(2n - 1) -
冉昭莘19539992020 ______ 2n!=1*2*3*4*...*2n n!=1*2*3*4*...*n 2^n=2*2*2*2*2*..*2(n个2) 所以2^n*n!=1*2*3*...*n*2*2*2*..*2(n个2) =(1*2)*(2*2)*(3*2)*...*(n*2) =2*4*6*8*...*2n 所以2n!/(n!*2^n)=1*2*3*4*...*2n/(2*4*6*8*...*2n) =1*3*5*7*...*(2n-1)

湛尚府925证明2n+1的阶乘除以2n的阶乘大于根号(n+1)1*3*5*7*……(2n+1)/[2*4……2n]大于根号(n+1) -
冉昭莘19539992020 ______[答案] 1,设等差数列(An)的公差d不为0,a1=9d,若ak是a1与a2k的等比中项,1 A 2 D 根据公式就可以相互的运算的 1) a1=9d,a2=a1+d

湛尚府9252(n+1)的阶乘是怎么算的? -
冉昭莘19539992020 ______ 2n加1

湛尚府925证明:2N的阶乘除以2的N次成以N的阶乘=1.3.5.7.(2n - 1) -
冉昭莘19539992020 ______[答案] 答: (2n)! =1*2*3*...*(2n-1)*2n =1*3*5*...*(2n-1)*2*4*6*...*2n =1*3*5*...*(2n-1)*2^n*(1*2*3*..*n) =1*3*5*...*(2n-1)*2^n*n! 所以(2n)!/(2^n*n!) =1*3*5*...*(2n-1)*2^n*n!/(2^n*n!) =1*3*5*...*(2n-1)

湛尚府925n乘以n加1的和除以2 与 n乘以n减1的差除以2问题规律 -
冉昭莘19539992020 ______ n乘以n加1的和除以2,是从1加到n的和, n乘以n减1的差除以2,是从1加到n-1的和

湛尚府925求(n+1)的阶乘/(2n+1)的双阶乘的极限 -
冉昭莘19539992020 ______[答案] 1、本题看山去是无穷大比无穷大型不定式;2、写出分子分母的通项后发现,是无穷个小于1的项相乘,结果为0. 解答如下:

湛尚府925请大家找出这样的n 1*2*3*4……*n+1=b^2 即n!+1=b^2 即n的阶乘加一,是一个 -
冉昭莘19539992020 ______ -module(factorialp1).-compile(export_all).fac(0) -> io:format(＂N=0,B=1~n＂), 1; fac(N) when N > 0 -> Temp = N*fac(N-1), %io:format(＂~p~n＂, [Temp]), Rn = math:sqrt(Temp+1), case Rn == erlang:trunc(Rn) of true -> io:format(＂N=~p,B=~p~n＂, ...

湛尚府925阶乘的问题.1)[(n+1)!]^2/[2(n+1)]!2) (2n)!/(n!)^2 3)(2n+1)!/[2(n+1)+1]!此类阶乘间的约分是如何约的,或者说有规律吗? -
冉昭莘19539992020 ______[答案] 没有什么公式,就是利用阶乘的定义和组合的定义[(n+1)!]^2/[2(n+1)]!=[(n+1)!][(n+1)!]/{(2n+2)(2n+1)(2n)……(n+2)[(n+1)!]}=[(n+1)!]/{(2n+2)(2n+1)(2n)……(n+2)}=1/C{(2n+2),(n+1)}2) (2n)!/(n!)^2=C{2n,n}...

湛尚府925阶乘化简1/2+1/6+1/12+……+1/n(n–1)
冉昭莘19539992020 ______ 1/2+1/6+1/12+……+1/n(n–1)=1-1/2+1/2-1/3+1/3-1/4+......+1/n-1/(n+1)=1-1/(n+1)=n/(n+1)