Răspuns:
S1=(1 1² k(1+k))
(2k-!) 0 3k-2)
------------------------
Sn=(n n² n(n+1)
(2n-1 0 3n-2)
_______________-------
a11=∑k= 1+2+...+n=n(n+1)/2
a12=∑k²=1²+2²+....+n²=n(n+1)(2n+1)/6
a13=∑(k²+k)=∑k²+∑k
∑k²=n(n+1)(2n+1)/6
∑k=n(n+1)/2
a13=n(n+1)(2n+1)/6+n(n+1)/2
randul 2
a21=∑(2k-1)=1+3+5+...+n(2n-1)=n²
a22=∑0=0
a32=∑(3k-2)=
∑3k-∑(-2)=3∑k-2n=
3n(n+1)/2-2n
Explicație pas cu pas:
