Răspuns:
∑Ak=( ∑(-1/2)^k ∑1/(4k²-1))
(∑1/(k²+3k+2) ∑k(k+1) 0
a11=∑(-1/2)^k=1-1/2+1/2²-1/2³+...+(1/2)ⁿ=progresie geometrica=1*[(-1/2)ⁿ-1]/(1-(-1/2)=2[(-1/2)ⁿ-1]=
a12=∑1/4k²-1)=∑1/[(2k-1)(2k+1)=
1/1*3+1/3*5+...1/(2n-1`)(2n+1)=
1/2[1-1/3+1/3-1/5+...+1/2n-1)-1/(2n+1)=1/2(1-1/2n+1)=1/2(2n+1-1)/2n+1=n/(2n+1)
a21=∑1/(k²+3k+2)=∑1/(k+1)(k+2)=1/(2*3+1/3*4+...+1/(n+1)(n+2)=
(1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2)=1/2-1/(n+2)=(n+2-1)/(n+2)=(n+1)/(n+2)
a22=∑(k²+k)=∑k²+∑k=
(1²+2²+3²+...n²)+(1+2+3+...+n)=
n(n+1)(2n+1)/6+n(n+1)/2=
n(n+1)[(2n+1)/6+1/2]=
n(n+1)(6n+3)/6
La sfarsit S= matricea formata din (a11 a12 )
(a21 a22)
Explicație pas cu pas:
