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Question Number 45932 by Tawa1 last updated on 18/Oct/18

Show that: ((1.2^2 + 2.3^2 + ... + n(n + 1)^2 )/(1^2 .2 + 2^2 .3 + ... + n^2 (n + 1))) = ((3n + 5)/(3n + 1))

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\frac{\mathrm{1}.\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{2}.\mathrm{3}^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}^{\mathrm{2}} .\mathrm{2}\:+\:\mathrm{2}^{\mathrm{2}} .\mathrm{3}\:+\:...\:+\:\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}\:+\:\mathrm{1}\right)}\:\:=\:\:\frac{\mathrm{3n}\:+\:\mathrm{5}}{\mathrm{3n}\:+\:\mathrm{1}} \\ $$

Commented by math khazana by abdo last updated on 19/Oct/18

S_n =((Σ_(k=1) ^n k(k+1)^2 )/(Σ_(k=1) ^n k^2 (k+1))) =((Σ_(k=1) ^n k(k^2 +2k+1))/(Σ_(k=1) ^n k^3 +Σ_k ^n k^2 )) =((Σ_(k=1) ^n k^3 +2Σ_(k=1) ^n k^2 +Σ_(k=1) ^n k)/(Σ_(k=1) ^n k^3 +Σ_(k=1) ^n k^2 )) =((((n^2 (n+1)^2 )/4)+2 ((n(n+1)(2n+1))/6)+((n(n+1))/2))/(((n^2 (n+1)^2 )/4)+((n(n+1)(2n+1))/6))) =((((n(n+1))/2) +(2/3)(2n+1) +1)/(((n(n+1))/2) +((2n+1)/3))) =((3n(n+1)+4(2n+1)+6)/(3n(n+1)+4n+2)) =((3n^2 +3n+8n+4+6)/(3n^2 +3n+4n+2)) =((3n^2 +11n+10)/(3n^2 +7n +2)) roots of 3n^2 +7n+2 Δ=49−4.3.2=49−24=25⇒n_1 =((−7+5)/6) =−(1/3) n_2 =((−7−5)/6) =−2 roots of 3n^2 +11n+10→Δ=121−4.3.10 =121−120=1 ⇒n_3 =((−11+1)/6) =−(5/3) n_3 =((−11−1)/6) =−2 ⇒ S_n =((3(n+(1/3))(n+2))/(3(n+(5/3))(n+2))) =((3n+1)/(3n+5)) and there is a error at the Q.

$${S}_{{n}} =\frac{\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)}\:=\frac{\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)}{\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \:+\sum_{{k}} ^{{n}} {k}^{\mathrm{2}} } \\ $$$$ \\ $$$$=\frac{\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \:+\sum_{{k}=\mathrm{1}} ^{{n}} {k}}{\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} +\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} } \\ $$$$=\frac{\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}}{\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}} \\ $$$$=\frac{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}{n}+\mathrm{1}\right)\:+\mathrm{1}}{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)+\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{6}}{\mathrm{3}{n}\left({n}+\mathrm{1}\right)+\mathrm{4}{n}+\mathrm{2}} \\ $$$$=\frac{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{8}{n}+\mathrm{4}+\mathrm{6}}{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{4}{n}+\mathrm{2}}\:=\frac{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{11}{n}+\mathrm{10}}{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{7}{n}\:+\mathrm{2}} \\ $$$${roots}\:{of}\:\mathrm{3}{n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{2} \\ $$$$\Delta=\mathrm{49}−\mathrm{4}.\mathrm{3}.\mathrm{2}=\mathrm{49}−\mathrm{24}=\mathrm{25}\Rightarrow{n}_{\mathrm{1}} =\frac{−\mathrm{7}+\mathrm{5}}{\mathrm{6}}\:=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${n}_{\mathrm{2}} =\frac{−\mathrm{7}−\mathrm{5}}{\mathrm{6}}\:=−\mathrm{2} \\ $$$${roots}\:{of}\:\mathrm{3}{n}^{\mathrm{2}} +\mathrm{11}{n}+\mathrm{10}\rightarrow\Delta=\mathrm{121}−\mathrm{4}.\mathrm{3}.\mathrm{10} \\ $$$$=\mathrm{121}−\mathrm{120}=\mathrm{1}\:\Rightarrow{n}_{\mathrm{3}} =\frac{−\mathrm{11}+\mathrm{1}}{\mathrm{6}}\:=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${n}_{\mathrm{3}} =\frac{−\mathrm{11}−\mathrm{1}}{\mathrm{6}}\:=−\mathrm{2}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{\mathrm{3}\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}\left({n}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left({n}+\mathrm{2}\right)}\:=\frac{\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}{n}+\mathrm{5}}\:\:{and}\:{there}\:{is}\:{a}\:{error} \\ $$$${at}\:{the}\:{Q}. \\ $$

Commented by math khazana by abdo last updated on 19/Oct/18

error at the final line S_n =((3(n+(5/3))(n+2))/(3(n+(1/3))(n+2))) =((3n+5)/(3n+1)) and ther is no error at the Q...

$${error}\:{at}\:{the}\:{final}\:{line}\:{S}_{{n}} =\frac{\mathrm{3}\left({n}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{3}{n}+\mathrm{5}}{\mathrm{3}{n}+\mathrm{1}}\:{and}\:{ther}\:{is}\:{no}\:{error}\:{at}\:{the}\:{Q}... \\ $$

Commented by Tawa1 last updated on 19/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 19/Oct/18

you are welcome sir

$${you}\:{are}\:{welcome}\:{sir} \\ $$

Answered by math1967 last updated on 19/Oct/18

S_n =1.2^2 +2.3^2 +.....+n(n+1)^2 =t_1 +t_(2 ) .......+n^3 +2n^2 +n ∴t_(1 ) +t_2 +......t_n =(1^3 +2^3 +...n^3 )+2(1^2 +2^2 +..+n^2 )+ (1+2+..n) ∴S_(n ) =((n^2 (n+1)^2 )/4) +2((n(n+1)(2n+1))/6)+((n(n+1))/2) =((n(n+1)(3n+5)(n+2))/(12)) Again 1^2 .2+2^2 .3+...n^2 (n+1) ∴t_n =n^3 +n^2 ∴S_n =(1^3 +2^3 +....+n^3 ) +(1^2 +2^2 +..+n^2 ) =((n^2 (n+1)^2 )/4)+((n(n+1)(2n+1))/6) =((n(n+1)(n+2)(3n+1))/(12)) ∴((1.2^2 +2.3^2 +....n(n+1)^2 )/(1^2 .2+2^2 .3+.....n^2 (n+1))) ((n(n+1)(3n+5)(n+2))/(12))×((12)/(n(n+1)(n+2)(3n+1))) =((3n+5)/(3n+1)) (proved)

$${S}_{{n}} =\mathrm{1}.\mathrm{2}^{\mathrm{2}} +\mathrm{2}.\mathrm{3}^{\mathrm{2}} +.....+{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$={t}_{\mathrm{1}} +{t}_{\mathrm{2}\:} .......+{n}^{\mathrm{3}} +\mathrm{2}{n}^{\mathrm{2}} +{n} \\ $$$$\therefore{t}_{\mathrm{1}\:\:} +{t}_{\mathrm{2}} +......{t}_{{n}} =\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +...{n}^{\mathrm{3}} \right)+\mathrm{2}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +..+{n}^{\mathrm{2}} \right)+ \\ $$$$\left(\mathrm{1}+\mathrm{2}+..{n}\right) \\ $$$$\therefore{S}_{{n}\:\:\:} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:+\mathrm{2}\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{5}\right)\left({n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$$${Again}\:\mathrm{1}^{\mathrm{2}} .\mathrm{2}+\mathrm{2}^{\mathrm{2}} .\mathrm{3}+...{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right) \\ $$$$\therefore{t}_{{n}} ={n}^{\mathrm{3}} +{n}^{\mathrm{2}} \\ $$$$\therefore{S}_{{n}} =\left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +....+{n}^{\mathrm{3}} \right)\:+\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +..+{n}^{\mathrm{2}} \right) \\ $$$$=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\: \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$$$\therefore\frac{\mathrm{1}.\mathrm{2}^{\mathrm{2}} +\mathrm{2}.\mathrm{3}^{\mathrm{2}} +....{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}^{\mathrm{2}} .\mathrm{2}+\mathrm{2}^{\mathrm{2}} .\mathrm{3}+.....{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{5}\right)\left({n}+\mathrm{2}\right)}{\mathrm{12}}×\frac{\mathrm{12}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}{n}+\mathrm{5}}{\mathrm{3}{n}+\mathrm{1}}\:\:\left({proved}\right) \\ $$

Commented by Tawa1 last updated on 19/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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