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Question Number 77953 by pete last updated on 12/Jan/20

Given that Σ_(r=0) ^4 6r =2Σ_(r=1) ^n 5r, work out the value of n.

$$\mathrm{Given}\:\mathrm{that}\:\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{6r}\:=\mathrm{2}\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{5r},\:\mathrm{work}\:\mathrm{out}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{n}. \\ $$

Answered by john santu last updated on 12/Jan/20

Σ_0 ^4 ((3/5))r = Σ_(r=1) ^n r (3/5)(0+1+2+3+4)=(n/2)(n+1) 12 = n^2 +n ⇔ n^2 −n−12=0 n = 4 , n = −3 (no solution)

$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{5}}\right){r}\:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}}\left(\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)=\frac{{n}}{\mathrm{2}}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{12}\:=\:{n}^{\mathrm{2}} +{n}\:\Leftrightarrow\:{n}^{\mathrm{2}} −{n}−\mathrm{12}=\mathrm{0} \\ $$$${n}\:=\:\mathrm{4}\:,\:{n}\:=\:−\mathrm{3}\:\left({no}\:{solution}\right) \\ $$

Commented by pete last updated on 12/Jan/20

thanks but i dont understand please. kindly do some explaining for me.

$$\mathrm{thanks}\:\mathrm{but}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{understand}\:\mathrm{please}. \\ $$$$\mathrm{kindly}\:\mathrm{do}\:\mathrm{some}\:\mathrm{explaining}\:\mathrm{for}\:\mathrm{me}. \\ $$

Answered by JDamian last updated on 12/Jan/20

3Σ_(r=0) ^4 r =5Σ_(r=1) ^n r 3((0+4)/2)5=5((n+1)/2)n 12=n(n+1) n^2 +n−12=0

$$\mathrm{3}\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{r}\:=\mathrm{5}\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{r} \\ $$$$\mathrm{3}\frac{\mathrm{0}+\mathrm{4}}{\mathrm{2}}\mathrm{5}=\mathrm{5}\frac{{n}+\mathrm{1}}{\mathrm{2}}{n} \\ $$$$\mathrm{12}={n}\left({n}+\mathrm{1}\right) \\ $$$${n}^{\mathrm{2}} +{n}−\mathrm{12}=\mathrm{0} \\ $$

Commented by pete last updated on 12/Jan/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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