Question

Exercitiul A1 va rog

Answer 1

Răspuns:

S1=(1 1² k(1+k))

(2k-!) 0 3k-2)

------------------------

Sn=(n n² n(n+1)

(2n-1 0 3n-2)

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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: