Răspuns:
∑₁ⁿ(k+3)(k-2)=∑(k²+3k-2k-6)=∑(k²+k-6)=∑k²+∑k-∑₁ⁿ6
∑k²=1²+2²+...+n²=n(n+1)(2n+1)/6
∑k=1+2+...+n=n(n+1)/2
∑6=n*6=6n
Suma devine
n(n+1)(2n+1)/6+n(n+1)/2-6n=
n(n+1)(2n+1)/6+3n(n+1)/6-6*6n=
n[(n+1)(2n+1)+3n+3-36]/6
n(2n²+2n+n+1+3n+3-36]/6
n(2n²+6n-32)/6
Explicație pas cu pas:
