Răspuns:
∑(1 k)
(k³ k(k+1))
a11= 1+1+...+1 = n ori
a12=1+2+...+n=n(n+1)/2
a21=1³+2³+...+n³=[n(n+1)]²/4
a22=∑k(k+1)=∑k²+∑k
∑k²=1²+2²+,,,+n²=n(n+1)(2n+1)/6
A=(n n(n+1)/2)
[n(n+1)]²/4 n(n+1)(2n+1)/6)
b)A=∑(1 2^k*3^k+1)
(2^k 2^k*3^(-k))
a11=1+1=...+1=n
∑2^k*3^k+1=3∑6^k=3(6+6²+6^3+...+6^n)=ai o progresie geometrica de ratie 6=suma acestei progresii=
3*(6*(6^n-1)/(6-1)=3*6(6^n-1)/5=18(6^n-1)/5
a21=2^1+2^2+...+2^n=progresie geometrica de ratie 2=
2*(2^n-1)/(2-1)=2*(2^n-1)
2^k*3^-k=(2/3)^k
a22=∑(2/3)^k=(2/3)+(2/3)^2+....+(2/3)6^n=progresie geometrica de ratiie (2/3)=
(2/3*[(2/3)^n-1]/(1-2/3)=2/3*[(2/3)^n-1]/(1/3)=
2[(2/3)^n-1]
construiesti matricea A cu termenii a11 a12,a21 a22 calculati mai sus
Explicație pas cu pas:
