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Question Number 34142 by Rio Mike last updated on 01/May/18
what is the derivative of (x+3)(x^2 + 5) and find the n sequence of Σ_(r=n+1 ) ^(2n) (4r^3 −3)
$${what}\:{is}\:{the}\:{derivative}\:{of}\: \\ $$$$\:\:\:\:\:\left(\boldsymbol{{x}}+\mathrm{3}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\:\mathrm{5}\right) \\ $$$${and}\:{find}\:{the}\:{n}\:{sequence}\:{of}\: \\ $$$$\:\:\:\underset{{r}={n}+\mathrm{1}\:} {\overset{\mathrm{2}{n}} {\sum}}\:\:\left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right) \\ $$
Commented by math khazana by abdo last updated on 01/May/18
(d/dx)( (x+3)(x^2 +5))= x^2 +5 +2x(x+3) = 3x^2 +6x +5 Σ_(r=n+1) ^(2n) (4r^3 −3) = 4Σ_(r=n+1) ^(2n) r^3 −3 Σ_(r=n+1) ^(2n) 1 Σ_(r=n+1) ^(2n) 1=2n−(n+1) +1=2n−n −1+1=n changement of indice r=n+p give A_n =Σ_(r=n+1) ^(2n) r^3 = Σ_(p=1) ^n (n+p)^3 =Σ_(p=1) ^n (n^3 +3n^2 p +3np^2 +p^3 ) =n^3 Σ_(p=1) ^n 1 +3n^2 Σ_(p=1) ^n p +3n Σ_(p=1) ^n p^2 +Σ_(p=1) ^n p^3 =n^4 +3n^2 ((n(n+1))/2) +3n ((n(n+1)(2n+1))/6) + ((n^2 (n+1)^2 )/4) A_n =((3n^3 (n+1))/2) +((n^2 (n+1)(2n+1))/2) +((n^2 (n+1)^2 )/4) ⇒ Σ_(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 .
$$\frac{{d}}{{dx}}\left(\:\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{5}\right)\right)=\:{x}^{\mathrm{2}} +\mathrm{5}\:\:+\mathrm{2}{x}\left({x}+\mathrm{3}\right) \\ $$$$=\:\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{6}{x}\:+\mathrm{5} \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right)\:=\:\mathrm{4}\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:{r}^{\mathrm{3}} \:−\mathrm{3}\:\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\mathrm{1} \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \mathrm{1}=\mathrm{2}{n}−\left({n}+\mathrm{1}\right)\:+\mathrm{1}=\mathrm{2}{n}−{n}\:−\mathrm{1}+\mathrm{1}={n} \\ $$$${changement}\:{of}\:{indice}\:{r}={n}+{p}\:{give} \\ $$$${A}_{{n}} =\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} {r}^{\mathrm{3}} \:=\:\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({n}+{p}\right)^{\mathrm{3}} \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({n}^{\mathrm{3}} \:+\mathrm{3}{n}^{\mathrm{2}} {p}\:+\mathrm{3}{np}^{\mathrm{2}} \:+{p}^{\mathrm{3}} \right) \\ $$$$={n}^{\mathrm{3}} \sum_{{p}=\mathrm{1}} ^{{n}} \mathrm{1}\:\:+\mathrm{3}{n}^{\mathrm{2}} \:\sum_{{p}=\mathrm{1}} ^{{n}} {p}\:\:+\mathrm{3}{n}\:\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{2}} \:\:+\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{3}} \\ $$$$={n}^{\mathrm{4}} \:\:+\mathrm{3}{n}^{\mathrm{2}} \:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{3}{n}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${A}_{{n}} =\frac{\mathrm{3}{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right)=\:\mathrm{4}{A}_{{n}} \:−\mathrm{3}{n}\:. \\ $$$$=\mathrm{6}{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)\:+\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\:+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{3}{n}\:. \\ $$

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