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Question Number 45155 by rahul 19 last updated on 09/Oct/18

Commented by rahul 19 last updated on 09/Oct/18

It′s a very beautiful Ques....

$${It}'{s}\:{a}\:{very}\:{beautiful}\:{Ques}.... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Oct/18

S_n =Σ_n ^n (r^4 /n^5 )+((r^3 n)/n^(5 ) )+((r^2 n^2 )/n^5 )+((2n^4 )/n^5 ) =(1/n^5 )Σ_(r=1) ^n r^4 +(1/n^4 )Σ_(r=1) ^n r^3 +(1/n^3 )Σ_(r=1) ^n r^2 +(2/n)Σ_(r=1) ^n 1 =(1/n^5 )((n^5 /5)+(n^4 /2)+(n^3 /3)−(n/(30)))+(1/n^4 )((n^4 /4)+(n^3 /2)+(n^2 /4))+(1/n^3 )((n^3 /3)+(n^2 /2)+(n/6))+(2/n)(n) =(1/5)+(1/(2n))+(1/(3n^2 ))−(1/(30n^4 ))+(1/4)+(1/(2n))+(1/(4n^2 ))+(1/3)+(1/(2n))+(1/(6n^2 ))+2 =((1/5)+(1/4)+(1/3)+2)+f(n) =((167)/(60))+f(n) S_n >((167)/(60))

$${S}_{{n}} =\underset{{n}} {\overset{{n}} {\sum}}\frac{{r}^{\mathrm{4}} }{{n}^{\mathrm{5}} }+\frac{{r}^{\mathrm{3}} {n}}{{n}^{\mathrm{5}\:} }+\frac{{r}^{\mathrm{2}} {n}^{\mathrm{2}} }{{n}^{\mathrm{5}} }+\frac{\mathrm{2}{n}^{\mathrm{4}} }{{n}^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{4}} +\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{3}} +\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}} +\frac{\mathrm{2}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\left(\frac{{n}^{\mathrm{5}} }{\mathrm{5}}+\frac{{n}^{\mathrm{4}} }{\mathrm{2}}+\frac{{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}}{\mathrm{30}}\right)+\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\left(\frac{{n}^{\mathrm{4}} }{\mathrm{4}}+\frac{{n}^{\mathrm{3}} }{\mathrm{2}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\left(\frac{{n}^{\mathrm{3}} }{\mathrm{3}}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{{n}}{\mathrm{6}}\right)+\frac{\mathrm{2}}{{n}}\left({n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} }+\mathrm{2} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}\right)+{f}\left({n}\right) \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+{f}\left({n}\right) \\ $$$${S}_{{n}} >\frac{\mathrm{167}}{\mathrm{60}} \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Oct/18

T_n =(1/n^5 )Σ_(r=0) ^(n−1) r^4 +(1/n^4 )Σ_(r=0) ^(n−1) r^3 +(1/n^3 )Σ_(r=0) ^(n−1) r^2 +(2/n)Σ_(r=0) ^(n−1) 1 ={(1/n^5 )Σ_(r=1) ^n r^4 +(1/n^4 )Σ_(r=1) ^n r^3 +(1/n^3 )Σ_(r=1) ^n r^2 +(2/n)Σ_(r=1) ^n }−{(n^4 /n^5 )+(n^3 /n^4 )+(n^2 /n^3 )} =((167)/(60))+f(n)−{(1/n)+(1/n)+(1/n)} =((167)/(60))+((3/(2n))+(3/(4n^2 ))−(1/(30n^4 )))−(3/n) =((167)/(60))+((3/(4n^2 ))−(1/(30n^4 ))−(3/(2n))) =((167)/(60))+(((45n^2 −2−90n^3 )/(60n^4 ))) =((167)/(60))+((45n^2 (1−n)−2)/(60n^4 )) =((167)/(60))−{((45n^2 (n−1)+2)/(60n^4 ))} so T_n <((167)/(60)) value of f(n) found in S_n f(n)=(1/(2n))+(1/(3n^2 ))−(1/(30n^4 ))+(1/(2n))+(1/(4n^2 ))+(1/(2n))+(1/(6n^2 )) =((3/(2n))+(9/(12n^2 ))−(1/(30n^4 )))

$${T}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{r}^{\mathrm{4}} +\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{r}^{\mathrm{3}} +\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{r}^{\mathrm{2}} +\frac{\mathrm{2}}{{n}}\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{1} \\ $$$$\:\:=\left\{\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{4}} +\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{3}} +\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}} +\frac{\mathrm{2}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\right\}−\left\{\frac{{n}^{\mathrm{4}} }{{n}^{\mathrm{5}} }+\frac{{n}^{\mathrm{3}} }{{n}^{\mathrm{4}} }+\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{3}} }\right\} \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+{f}\left({n}\right)−\left\{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}\right\} \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+\left(\frac{\mathrm{3}}{\mathrm{2}{n}}+\frac{\mathrm{3}}{\mathrm{4}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }\right)−\frac{\mathrm{3}}{{n}} \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+\left(\frac{\mathrm{3}}{\mathrm{4}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{2}{n}}\right) \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+\left(\frac{\mathrm{45}{n}^{\mathrm{2}} −\mathrm{2}−\mathrm{90}{n}^{\mathrm{3}} }{\mathrm{60}{n}^{\mathrm{4}} }\right) \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}+\frac{\mathrm{45}{n}^{\mathrm{2}} \left(\mathrm{1}−{n}\right)−\mathrm{2}}{\mathrm{60}{n}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{167}}{\mathrm{60}}−\left\{\frac{\mathrm{45}{n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)+\mathrm{2}}{\mathrm{60}{n}^{\mathrm{4}} }\right\} \\ $$$${so}\:{T}_{{n}} <\frac{\mathrm{167}}{\mathrm{60}} \\ $$$${value}\:{of}\:{f}\left({n}\right)\:{found}\:{in}\:{S}_{{n}} \\ $$$${f}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} } \\ $$$$\:\:\:=\left(\frac{\mathrm{3}}{\mathrm{2}{n}}+\frac{\mathrm{9}}{\mathrm{12}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{30}{n}^{\mathrm{4}} }\right) \\ $$

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