Wyznacz: 1*1!+2*2!+...+n*n!=
1*1!+2*2!+...+n*n!=
dla n=1
1*1=1
dla n=2
1+2*2=5=6-1=3!-1=(2+1)!-1
dla n=3
5+3*3!=23 =24-1=4!-1=(3+1)!-1
...
1*1!+2*2!+...+n*n!=(n+1)!-1
Dowod indukcyjny:
1. Spr. dla n=1
1=2-1=1
L=P
Zalozenie:
Teza
dla n+1
(1*1!+2*2!+...+n*n!-1)+(n+1)*(n+1)!=(n+2)!-1
Dowod:
(1*1!+2*2!+...+n*n!-1)+(n+1)*(n+1)!=(n+1)!-1+(n+1)(n+1)!=(n+1)!-1+(n+2-1)(n+1)!=
(n+1)!-1+(n+2)(n+1)!-(n+1)!=(n+2)!-1
c.n.d.
Dla przykładu:
1! = 1 * 1 =1
2 = 1 + 2 * 2 = 5 = 6 - 1 = 3! - 1 = ( 2 + 1 )! - 1
n! = n * ( n - 1 )!
1 * 1! + 2 * 2! + ... + n * n! = ( n + 1 )! - 1
Spr.: dla n = 1
1 = ( n + 1 )! - 1 = ( 1 + 1 )! -1 = 2! - 1 = 1 * 2 - 1 = 2 - 1 = 1
Spr.: dla n = 2
2 = ( n + 1 )! -1 = ( 2 + 1 )! - 1 = 3! -1 = 1 * 2 * 3 - 1 = 6 - 1 = 5
dla n + 1
(1 * 1! + 2 * 2! + ... + n * n! - 1 ) + ( n + 1 ) * ( n + 1 )! = ( n + 2 )! - 1
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1*1!+2*2!+...+n*n!=
dla n=1
1*1=1
dla n=2
1+2*2=5=6-1=3!-1=(2+1)!-1
dla n=3
5+3*3!=23 =24-1=4!-1=(3+1)!-1
...
1*1!+2*2!+...+n*n!=(n+1)!-1
Dowod indukcyjny:
1. Spr. dla n=1
1=2-1=1
L=P
Zalozenie:
1*1!+2*2!+...+n*n!=(n+1)!-1
Teza
dla n+1
(1*1!+2*2!+...+n*n!-1)+(n+1)*(n+1)!=(n+2)!-1
Dowod:
(1*1!+2*2!+...+n*n!-1)+(n+1)*(n+1)!=(n+1)!-1+(n+1)(n+1)!=(n+1)!-1+(n+2-1)(n+1)!=
(n+1)!-1+(n+2)(n+1)!-(n+1)!=(n+2)!-1
c.n.d.
Dla przykładu:
1! = 1 * 1 =1
2 = 1 + 2 * 2 = 5 = 6 - 1 = 3! - 1 = ( 2 + 1 )! - 1
n! = n * ( n - 1 )!
1 * 1! + 2 * 2! + ... + n * n! = ( n + 1 )! - 1
Spr.: dla n = 1
1 = ( n + 1 )! - 1 = ( 1 + 1 )! -1 = 2! - 1 = 1 * 2 - 1 = 2 - 1 = 1
Spr.: dla n = 2
2 = ( n + 1 )! -1 = ( 2 + 1 )! - 1 = 3! -1 = 1 * 2 * 3 - 1 = 6 - 1 = 5
dla n + 1
(1 * 1! + 2 * 2! + ... + n * n! - 1 ) + ( n + 1 ) * ( n + 1 )! = ( n + 2 )! - 1