Question

Sa se calculeze suma : (atasament) . Multumesc mult !!!

Answer 1

Se simplifica ce e la numitor cu ce este la numarator si se obtine
S=2*3+3*4+4*5+..+(n+1)(n+2)[tex]S=2*3+3*4+4*5...+n^2+3n+2\\= (1^1+3*1+2)+(2^2+3*2+2)+...+(n^2+3n+2)=\\ =1^2+2^2+3^3+...+n^2+3(1+2+...+n)+(2+2+2+...+2)=\\ \frac{n(n+1)(2n+1)}{6}+3 \frac{n(n+1)}{2}+2n=\\ = \frac{n(2n^2+3n+1)}{6}+ \frac{9(n^2+n)}{6}+ \frac{12n}{6} =\\ =\frac{2n^3+3n^2+n+9n^2+9n+12n}{6}=\\ =\frac{2n^3+12n^2+22n}{6}= \\ =\frac{n^3+6n^2+11n}{3} [/tex]

Answer 2

$\displaystyle \mathtt{ \frac{3!}{1!}+ \frac{4!}{2!}+ \frac{5!}{3!}+...+ \frac{(n+2)!}{n!}=\sum\limits_{k=1}^n \frac{(k+2)!}{k!}=\sum\limits_{k=1}^n \frac{k!(k+1)(k+2)}{k!} =}\\ \\ \mathtt{=\sum\limits_{k=1}^n(k+1)(k+2)=\sum\limits_{k=1}^n(k^2+2k+k+2)=\sum\limits_{k=1}^n(k^2+3k+2)=}\\ \\ \mathtt{=\sum\limits_{k=1}^nk^2+\sum\limits_{k=1}^n3k+\sum\limits_{k=1}^n2= \sum\limits_{k=1}^nk^2+3\sum\limits_{k=1}^nk+\sum\limits_{k=1}^n2=}\\ \\ \mathtt{= \frac{n(n+1)(2n+1)}{6} +3 \cdot \frac{n(n+1)}{2} +2n=}\\ \\ \mathtt{= \frac{n(n+1)(2n+1)+9n(n+1)+12n}{6}=}\\ \\ \mathtt{ =\frac{n(2n^2+3n+1)+9n^2+9n+12n}{6}=\frac{2n^3+3n^2+n+9n^2+9n+12n}{6} =}\\ \\ \mathtt{= \frac{2n^3+12n^2+22n}{6}= \frac{2(n^3+6n^2+11n)}{6} = \frac{n^3+6n^2+11n}{3} }$