Question

Folosim metoda inductiei matematice sa se demonstreze ca pentru orice n€N* au loc egalitatile:

a) 1+2+3+...+n= n(n+1)/2


b) 1+5+9+...+(4n-3)=n(2n-1)

TOATE SUBPUNCTELE DIN IMAGINE! URGENT!!!

Answer 1

1 + 2 + 3 + ... + n = [n(n+1)]/2

1. Etapa de verificare :

p(1) : 1 * 2/2 = 1 , adevarat

2. Etapa de demonstrație

p(k)(1) -> p(k+1)(1)

p(k) : 1 + 2 + 3 + ... + k = [k(k+1)]/2

p(k+1) : 1 + 2 + 3 + ... + k + k + 1 = [ (k+1)(k+2)]/2

[k(k+1)]/2 = [ (k+1)(k+2)]/2

(k² + k )/2 + k + 1 = (k² + 3k + 2)/2

k² + k + 2k + 2 = k² + 3k + 2

k² + 3k + 2 = k² + 3k + 2 Adevarat

1 + 5 + 9 + ... + (4n - 3) = n( 2n-1)

1. Verificarea:

p(1) : 1 * 1 = 1 , adevarat

2. Etapa de demonstrație

p(k)(1) -> p(k+1)(1)

p(k) : 1 + 5 + 9 + ... + (4k-3) = k( 2k - 1)

p(k + 1) : 1 + 5 + 9 + ... + ( 4k - 3) + (4k + 1) = (k+1)(2k +1)

k(2k-1) + (4k-3) = (k+1)(2k+1)

2k² - k + 4k + 1 = 2k² + 3k + 1

2k² + 3k + 1 = 2k² + 3k + 1

1² + 2² + .... + n² = [n(n+1)(2n+1)]/6

1. Etapa de verificare :

p(1) : 1 = 1 * 2 * 3 / 6 , 1 = 1 , Adevarat

2. Demonstrația:

p(k)(1) -> p(k+1)(1)

p(k) : 1² + 2² + ... + k² = [k(k+1)(2k+1)]/6

p(k+1) : 1² + 2² + ... + k² + (k+1)² = [(k+1)(k+2)(2k+3)]/6

[k(k+1)(2k+1)]/6 + (k+1)² = [(k+1)(k+2)(2k+3)]/6

(k² + k)(2k + 1)/6 + k² + 2k + 1 = (k²+3k+2)(2k+3)/6

2k³ + k² + 2k² + k + 6k² + 12k + 6 = 2k³ + 6k² + 4k + 3k² + 9k + 6

2k³ + 9k² + 13k + 6 = 2k³ + 9k² + 13k + 6