Question

1*3+2*5+3*7+.....+n(2n+1) sa se calculeze suma la mate

Answer 1

[tex]1\cdot 3+2\cdot 5+3\cdot 7+\ldots+n(2n+1)=\\ \displaystyle\sum\limits_{k=1}^{n}k(2k+1)=\displaystyle\sum\limits_{k=1}^{n}(2k^2+k)=2\displaystyle\sum\limits_{k=1}^{n}k ^2+\displaystyle\sum\limits_{k=1}^{n}k=\\ =2\cdot \dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(2n+1)}{3}+\dfrac{n(n+1)}{2}=\\ =\dfrac{2n(n+1)(2n+1)+3n(n+1)}{6}=\boxed{\dfrac{n(n+1)(4n+5)}{6}}[/tex]