Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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 .

$$\frac{{d}}{{dx}}\left(\:\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{5}\right)\right)=\:{x}^{\mathrm{2}} +\mathrm{5}\:\:+\mathrm{2}{x}\left({x}+\mathrm{3}\right) \\ $$$$=\:\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{6}{x}\:+\mathrm{5} \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right)\:=\:\mathrm{4}\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:{r}^{\mathrm{3}} \:−\mathrm{3}\:\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\mathrm{1} \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \mathrm{1}=\mathrm{2}{n}−\left({n}+\mathrm{1}\right)\:+\mathrm{1}=\mathrm{2}{n}−{n}\:−\mathrm{1}+\mathrm{1}={n} \\ $$$${changement}\:{of}\:{indice}\:{r}={n}+{p}\:{give} \\ $$$${A}_{{n}} =\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} {r}^{\mathrm{3}} \:=\:\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({n}+{p}\right)^{\mathrm{3}} \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({n}^{\mathrm{3}} \:+\mathrm{3}{n}^{\mathrm{2}} {p}\:+\mathrm{3}{np}^{\mathrm{2}} \:+{p}^{\mathrm{3}} \right) \\ $$$$={n}^{\mathrm{3}} \sum_{{p}=\mathrm{1}} ^{{n}} \mathrm{1}\:\:+\mathrm{3}{n}^{\mathrm{2}} \:\sum_{{p}=\mathrm{1}} ^{{n}} {p}\:\:+\mathrm{3}{n}\:\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{2}} \:\:+\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{3}} \\ $$$$={n}^{\mathrm{4}} \:\:+\mathrm{3}{n}^{\mathrm{2}} \:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\mathrm{3}{n}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${A}_{{n}} =\frac{\mathrm{3}{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow \\ $$$$\sum_{{r}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right)=\:\mathrm{4}{A}_{{n}} \:−\mathrm{3}{n}\:. \\ $$$$=\mathrm{6}{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)\:+\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\:+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{3}{n}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com