2+n+1+的阶乘

褚育常2355n平方的阶乘为多少1^2+...+n^2=? -
柏鹏琰13451201252 ______[答案] 1^2+...+n^2=n(n+1)(2n+1)/6

褚育常2355对 n\(n+1)!求和 注意分母是n+1的阶乘 -
柏鹏琰13451201252 ______[答案] 由k/(k+1)!=(k+1-1)/(k+1)!=1/k!-1/(k+1)!, 所以1/2!+2/3!+3/4!+…+n/(n+1)! =1/1!-1/2!+1/2!-1/3!+1/3!-1/4!+…+1/n!-1/(n+1)! =1-1/(n+1)!

褚育常2355阶乘问题一个求和:1*1!+2*2!+3*3!+…+n*n! -
柏鹏琰13451201252 ______[答案] 任何大于1的自然数n阶乘表示方法n!=1*2*3*……*n.例如4!,则阶乘式是1*2*3*4.求和1*1!+2*2!+3*3!+…+n*n!每个乘积是N乘以N的阶乘再相加~完毕~

褚育常2355关于阶乘 证明:1!+2*2!+3*3!+……+n*n!=(n+1)! - 1 -
柏鹏琰13451201252 ______ 1!+2*2!+3*3!+……+n*n!=(1+1)1!+(2+1)2!+...+(n+1)n!-1!-2!-...-n!=(n+1)!-1

褚育常2355证明(2n)的阶乘整除[(n)的阶乘乘以(n+1)的阶乘] -
柏鹏琰13451201252 ______[答案] 郭敦顒回答: [n!•(n+1)!]= n!•n!(n+1), (2n)!=n!•(n+1)(n+2)•…•(2n-1)•(2n) (n+1)(n+2)•…•(2n-1)•(2n)>n!(n+1) n!•(n+1)(n+2)•…•(2n-1)•(2n)>n!•n!(n+1) ∴(2n)!不能整除[n!•(n+1)!]. 但(2n)!能整除[n!•(n+1)]! ∵[n!•(n+1)]!=[1•...

褚育常2355一的阶乘加二的阶乘一直加到n的阶乘,这个和再除以n的阶乘得多少? -
柏鹏琰13451201252 ______[答案] Lim n->无穷1!+2!+3!+n!/n!=1+1/n+1/[n(n-1)]+1/[n(n-1)(n-2)]...+1/n!=1

褚育常2355计算1!+2!+3!...+(n - 1)!+n!.设计求解该问题的C语言程序,阶乘的计算使用递归函数实现 -
柏鹏琰13451201252 ______ #include fun(int x); int main(void) { char *pszBuff = NULL; int a; int i=1; int sumResult = 0; printf(＂input a int number::＂); scanf(＂%d＂, &a); for(i=1;i<=a;i++) { int tmp = fun(i); sumResult = sumResult + tmp ; printf(＂%d!=%d, 1-%d的阶乘的和为%d\n...

褚育常2355n的阶乘比上(n+1)的阶乘等于多少 -
柏鹏琰13451201252 ______[答案] n!/(n+1)! =(1*2*······*n)/(1*2*······*n*(n+1)) =1/(n+1)

褚育常2355关于阶乘 证明:+2*2!+3*3!+……+n*n!=(n+1)! - 1 -
柏鹏琰13451201252 ______[答案] 1!+2*2!+3*3!+……+n*n!=(1+1)1!+(2+1)2!+...+(n+1)n!-1!-2!-...-n!=(n+1)!-1

褚育常2355阶乘化简1/2+1/6+1/12+……+1/n(n–1)
柏鹏琰13451201252 ______ 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)