Question

Exercitiul A1 doar c si d

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Răspuns:

Vom demonstra folosind inductia matematica.

c) 1·2·3+3·4·5+...+(2n-1)2n(2n+1)=n(n+1)(2n^2+2n-1)

1) Verificarea: P(1) = 1*2*3 = 1(1+1)(2*1²+2*1-1) => 6=6 (adevarat)

2) Demonstratia

P(k):1·2·3+3·4·5+...+(2k-1)2k(2k+1)=k(k+1)(2k^2+2k-1)  

P(k+1):1·2·3+3·4·5+...+(2k-1)2k(2k+1)+(2k+1)(2k+2)(2k+3)=(k+1)(k+2)[2(k+1)^2+2(k+1)-1] <=>

<=> 1·2·3+3·4·5+...+(2k-1)2k(2k+1)+(2k+1)(2k+2)(2k+3)=(k+1)(k+2)(2k^2+6k+3)

P(k)+k+1 = k(k+1)(2k^2+2k-1)+(2k+1)(2k+2)(2k+3) =

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

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

(k+1)(2k^3+2k^2-k+8k^2+12k+4k+6)=

(k+1)(2k^3+10k^2+15k+6)=

(k+1)(k+2)(2k^2+6k+3)

d) 1*2^2+2*3^2+...+(n-1)n^2= $\frac{n(n^2-1)(3n+2)}{12}$

2) Demonstratia (Am trecut direct la demonstratie de data asta, dar in test pasul 1, adica verificarea este obligatorie, dar deja sti asta asa ca facem demonstratia)

P(k) = 1*2^2+2*3^2+...+(k-1)k^2=$\frac{k(k^2-1)(3k+2)}{12}$

P(k+1) = 1*2^2+2*3^2+...+(k-1)k^2+k(k+1)^2 =

=$\frac{(k+1)(k^2+2k)(3k+5)}{12}$ = $\frac{3k^4+14k^3+21k^2+10k}{12}$ (Am facut pe calculator calculele fara sa mai dau factor comun)

P(k)+k+1 = $\frac{k(k^2-1)(3k+2)}{12}$ + k(k+1)^2 = $\frac{3k^4+14k^3+21k^2+10k}{12}$(Am facut pe calculator calculele fara sa mai dau factor comun)

Explicație pas cu pas: