+2*2!+3*3!+4*4!…… n*n!

+2*2!+3*3!+4*4!…… n*n!

Let x = 1! + 2 * 2! + 3 * 3! + 4 * 4 N * n! Add 1! + 2! + 3! + 4!... on both sides N!
X + (1! + 2! + 3! + 4!...) N!
=2*1!+3*2!+4*3!+…… (n+1)n!=2!+3!+…… (n+1)!
So x + (1! + 2! + 3! + 4 n!)=2!+3!+…… (n+1)!
x=(n+1)!-1

In mathematics, there is an operation called factorial For example, 3! = 1 × 2 × 3, = 1 × 2 × 3 × 4, = 1 × 2 × 3 × 4 × 5, etc

36!/35!=36*(35!)/(35!)=36

For any positive integer n, define "double factorial n!" as follows: When n is even, = n · (n-2) · (n-4) 6·4·2 When n is odd, = n · (n-2) · (n-4) 5·3·1 Now there are four propositions （1）(2011!)·(2010!)=2011! （2）2010!=2×1005! (3) The single digit of 2010! Is 0 (4) The single digit of 2011! Is 5 The correct propositions are

1. Three and five are right
1. 2011! Is the product of all odd numbers below 2011 (including 2011), and 2010! Is the product of all even numbers below 2010 (including 2010). Multiplication together is the product of all positive integers below 2011 (including 2011). So it is 2011!
2. 2010! Is 1005 times the 2 of 1005
3. 2010! Has a factor of 10, so the bit must be 0
4. 2011! Is the multiplication of odd numbers, of which there are 5, so 2011! Is an odd multiple of 5, so the bit must be 5

What does factorial mean? I want to know the concrete expression of n

N! Is to multiply from 1 by a number one greater than the previous one, and multiply to n
Specifically: 1 * 2 * 3 * 4. * (n-2) * (n-1) * n = n!

Enter a positive integer n to calculate 1 + 1 / 2! + 1 / 3 The sum of 1 / N! Is output. The operation definition of computing factorial is required Use C language!

#include
void main()
{
int n,i=1,fa=1;
double sum=1;
scanf("%d",&n);
for (i=1;i

When n is greater than or equal to 6, it is proved that the factorial of n is greater than the third power of n

It is proved that: when n = 6! = 7206? = 216, so 6! > 6?, if n = k, the original formula holds, that is, K! > k? Then when n = K + 1, the left = (K + 1)! = (K + 1) * k! The right = (K + 1) 3 = (K + 1) * (K + 1) 2, because K! > k? And when k > 3, K? > (K + 1) 2

A proof that the sum of factorials of n is equal to e?

Using Taylor expansion: FX = f (a) + F '(a) / 1! (x-a) + F' '(a) / 2! (x-a) ^ 2 +
e^x=f(0)+f'(0)*x/1!+f''（0）x^2/2!+.
e=1+1/2!+1/3!+...1/n!

1*n+2*(n-1)+3*(n-2)+… +Proof of n * 1 = 1 / 6N (n + 1) (n + 2) mathematical induction As the title It is proved by mathematical induction. 1. When n = 1 2.… In this way

1. When n = 1, left = 1, right = (1 / 6) * 1 * (1 + 1) * (1 + 2) = 1, left = right,
So the original equation holds
2. If n = K (k > = 1), the original equation also holds,
That is, 1 * k + 2 * (k-1) + 3 * (K-2) +... + k * 1 = (1 / 6) K (K + 1) (K + 2) holds
3. When n = K + 1, the left side of the original equation = 1 * (K + 1) + 2 * [(K + 1) - 1] + 3 * [(K + 1) - 2] +... + (K + 1) * 1
=[1*k+1]+[2*(k-1)+2]+[3*(k-2)+3]+…… +[k*1+1]
=[1*k+2*（k-1)+3*(k-2)+...+k*1]+[1+2+3+…… +(k+1)]
=(1 / 6) K (K + 1) (K + 2) + (K + 1) (K + 2) / 2
=(1/6)(k+1)(k+2)(k+3)
On the right side = (1 / 6) (K + 1) [(K + 1) + 1] [(K + 1) + 2] = (1 / 6) (K + 1) (K + 2) (K + 3),
Left = right,
Therefore, when n = K + 1, the original equation also holds
5. To sum up, the original equation holds for any positive integer n

As the title It is proved by mathematical induction that 1 / N + 1 / (1 + n) + 1 / (n + 2) +. 1 / N ^ 2 > 1 (n ∈ N and N > 1) Therefore, when n = K + 1, there are: 1/n+1/(n+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1) >1+1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1) This step is wrong. We should start from 1 / (n + 1) and add should be > 1 + 1 / (k ^ 2 + 1) + 1 / (k ^ 2 + 2) + 1 / (k ^ 2 + 2K + 1) - 1 / n It is proved that 1 / (k ^ 2 + 1) + 1 / (k ^ 2 + 2) + 1 / (k ^ 2 + 2K + 1) - 1 / N > 0

prove:
(1) When n = 2,
1 / 2 + 1 / 3 + 1 / 4 = 13 / 12 > 1 holds
(2) Suppose that when n = k, i.e
1/k+1/(k+1)+...+1/k^2>1
Therefore, when n = K + 1, there are:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
=1/k+1/(k+1)+...+1/k^2+[1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)-1/k]
>1+[1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/k]
>1+[(2k+1)/(k^2+2k+1)-1/k]
=1+[(2k²+k-k²-2k-1)/(k²+2k+1)k]
=1+[(k²-k-1)/(k²+2k+1)k]
Because:
K? - k-1 > 0 (when k > 2)
(k²-k-1)/(k²+2k+1)k>0
So:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
>1+0
=1
So when n = K + 1, the original formula holds
To sum up, there are:
1/n+1/(n+1)+1/(n+2)+… +1 / N ^ 2 > 1 (n > 1 and N is an integer)

Mathematical induction proves: 1 * n + 2 (n-1) + 3 (n-2) + +(n-1)*2+n*1=(1/6)n(n+1)(n+2) When n = 1, left = 1 * 1 = 1 Right side = 1 * 3 * 1 Left = right, the equation holds! If n = k, it holds (k > 1) 1*k+2(k-1)+3(k-2)+… +(k-1)*2+k*1=(1/6)k(k+1)(k+2) When n = K + 1; left =1*(k+1)+2(k+1-1)+3(k+1-2)+… +(k+1-1)*2+(k+1)*1 =1*k+1*1+2(k-1)+2*1+… +How did k * 1 + [K + (K + 1)] come from? =[1*k+2(k-1)+… +(k-1)*2+k*1]+【1+2+3+… +K + (K + 1)] ← why is it an arithmetic sequence? Isn't it 1 + 2 + 3 + ···· + 3 + 2 + 1? =(1/6)k(k+1)(k+2)+1+2+3+… +k+(k+1) =(1/6)k(k+1)(k+2)+1/2*(k+1)*(k+2) =(1/6)(k+1)(k+2)(k+3) =(1/6)(k+1)[(k+1)+1][(k+1)+2] =No one

If n = K + 1, then it is the sum of K + 1. Therefore, 1 * (K + 1) + 2 (K + 1-1) + 3 (K + 1-2) + 1 + (k+1-1)*2+(k+1)*1=1*k+1*1+2(k-1)+2*1+3(k-2)+3*1+… +K * 1 + [K + (K + 1)] (where (K + 1-1) * 2 + (K + 1) * 1 = k * 2 + (K + 1) * 1 = k * 1 + [K + (K + 1)])