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Question Number 34142 by Rio Mike last updated on 01/May/18

$$whatisthederivativeof$$$$(\mathit{x}+3)({\mathit{x}}^2+5)$$$$andfindthensequenceof$$$$\underset{r=n+1}{\sum ^{2n}}(4r^3-3)$$

Commented by math khazana by abdo last updated on 01/May/18

$$\frac{d}{dx}\left((x+3)(x^2+5)\right)=x^2+5+2x(x+3)$$$$=3x^2+6x+5$$$$\sum _{r=n+1}^{2n}(4r^3-3)=4\sum _{r=n+1}^{2n}{r}^3-3\sum _{r=n+1}^{2n}1$$$$\sum _{r=n+1}^{2n}1=2n-(n+1)+1=2n-n-1+1=n$$$$changementofindicer=n+pgive$$$$A_n=\sum _{r=n+1}^{2n}{r}^3=\sum _{p=1}^n{(n+p)}^3$$$$=\sum _{p=1}^n(n^3+3n^{2}p+3{np}^2+p^3)$$$$=n^3\sum _{p=1}^{n}1+3n^2\sum _{p=1}^{n}p+3n\sum _{p=1}^n{p}^2+\sum _{p=1}^n{p}^3$$$$=n^4+3n^2\frac{n(n+1)}{2}+3n\frac{n(n+1)(2n+1)}{6}+\frac{n^2{(n+1)}^2}{4}$$$$A_n=\frac{3n^3(n+1)}{2}+\frac{n^2(n+1)(2n+1)}{2}+\frac{n^2{(n+1)}^2}{4}\Rightarrow$$$$\sum _{r=n+1}^{2n}(4r^3-3)=4A_n-3n.$$$$=6n^3(n+1)+2n^2(n+1)(2n+1)+n^2{(n+1)}^2-3n.$$

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