Question

c de n luate 1 + c de n luate cate 2 =15

Answer 1

Răspuns:

Explicație pas cu pas:

$C_n^1+C_n^2=15, ~~=>n~natural,~n\geq 2.\\\dfrac{n!}{1!*(n-1)!}+\dfrac{n!}{2!*(n-2)!}=15,~=>~\dfrac{(n-1)!*n}{1!*(n-1)!}+\dfrac{(n-2)!*(n-1)*n}{2!*(n-2)!}=15,~=>~n+\dfrac{(n-1)*n}{2}=15~|*2,~=>~2n+(n-1)*n=2*15$

⇒2n+n²-n-30=0, ⇒n²+n-30=0, Δ=1²-4·1·(-30)=1+120=121=11², deci

n=(-1-11)/2=-6, nu convine

n=(-1+11)/2=5.

Răspuns: n=5.

Answer 2

Răspuns:

Explicație pas cu pas:

n!/1!(n - 1)! + n!/2!(n - 2)! = 15

n(n - 1)! / (n - 1)! + n(n - 1)(n - 2)!/2(n - 2)! = 15

n + n(n - 1)/2 = 15

2n + n(n - 1) = 30

2n + n^2 - n - 30 = 0

n^2 + n - 30 = 0

Δ = 1 + 120 = 121

n = numar intreg pozitiv

n = (-1 + 11)/2 = 10/2 = 5