1-1-3-5-1-3-5-7-1-5-7-9-nth-term-

Question Number 46245 by rpseelam last updated on 23/Oct/18

$$\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{7}.\mathrm{9}}+…..{nth}\:{term} \\ $$

Commented by maxmathsup by imad last updated on 23/Oct/18

let S_n =Σ_(k=0) ^n (1/((2k+1)(2k+3)(2k+5))) let find S_n let decompose F(x)=(1/((2x+1)(2x+3)(2x+5))) ⇒F(x)=(a/(2x+1)) +(b/(2x+3)) +(c/(2x+5)) a =lim_(x→−(1/2)) (2x+1)F(x)= (1/(2.4)) =(1/8) b=lim_(x→−(3/2)) (2x+3)F(x)= (1/((−2).2)) =−(1/4) c=lim_(x→−(5/2)) (2x+5)F(x)= (1/((−4).(−2))) =(1/8) ⇒ F(x)= (1/(8(2x+1))) −(1/(4(2x+3))) +(1/(8(2x+5))) ⇒ S_n =(1/8)Σ_(k=0) ^n (1/(2k+1)) −(1/4) Σ_(k=0) ^n (1/(2k+3)) +(1/8) Σ_(k=0) ^n (1/(2k+5)) but Σ_(k=0) ^n (1/(2k+3)) =_(k=j−1) Σ_(j=1) ^(n+1) (1/(2j+1)) =Σ_(j=0) ^n (1/(2j+1)) −1 +(1/(2n+1)) Σ_(k=0) ^n (1/(2k+5)) =_(k=j−2) Σ_(j=2) ^(n+2) (1/(2j +1)) =Σ_(j=0) ^n (1/(2j+1)) −1−(1/3) +(1/(2n+1)) +(1/(2n+5)) S_n =(1/8) Σ_(k=0) ^n (1/(2k+1)) −(1/4) Σ_(j=0) ^n (1/(2j+1)) +(1/4) −(1/(4(2n+1))) +(1/8)Σ_(j=0) ^n (1/(2j+1)) +(1/8)(−(4/3)) +(1/(8(2n+1))) +(1/(8(2n+5))) =(1/4) −(1/6) −(1/(4(2n+1))) +(1/(8(2n+))) +(1/(8(2n+5))) =(1/(12)) −(1/(8(2n+1))) +(1/(8(2n+5))) =(1/(12)) +(1/8){(1/(2n+5)) −(1/(2n+1))} =(1/(12)) +(1/8){((−4)/((2n+1)(2n+5)))} =(1/(12)) −(1/(2(2n+1)(2n+5))) .

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)\left(\mathrm{2}{k}+\mathrm{5}\right)}\:\:\:{let}\:{find}\:\:{S}_{{n}} \:\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{5}\right)}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{3}}\:+\frac{{c}}{\mathrm{2}{x}+\mathrm{5}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left(\mathrm{2}{x}+\mathrm{1}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${b}={lim}_{{x}\rightarrow−\frac{\mathrm{3}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{x}+\mathrm{3}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(−\mathrm{2}\right).\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${c}={lim}_{{x}\rightarrow−\frac{\mathrm{5}}{\mathrm{2}}} \:\:\left(\mathrm{2}{x}+\mathrm{5}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(−\mathrm{4}\right).\left(−\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{3}\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{x}+\mathrm{5}\right)}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{5}}\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\:=_{{k}={j}−\mathrm{1}} \:\:\:\sum_{{j}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}{j}+\mathrm{1}}\:=\sum_{{j}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{j}+\mathrm{1}}\:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{5}}\:=_{{k}={j}−\mathrm{2}} \:\sum_{{j}=\mathrm{2}} ^{{n}+\mathrm{2}} \:\:\frac{\mathrm{1}}{\mathrm{2}{j}\:+\mathrm{1}}\:=\sum_{{j}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{j}+\mathrm{1}}\:−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{j}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{j}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$+\frac{\mathrm{1}}{\mathrm{8}}\sum_{{j}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{j}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left(−\frac{\mathrm{4}}{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:\:−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{n}+\mathrm{5}\right)}\:=\frac{\mathrm{1}}{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\frac{−\mathrm{4}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)}\right\}\:=\frac{\mathrm{1}}{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

T_n =(1/({1+(n−1)2}{3+(n−1)2}{5+(n−1)2})) T_n =(1/((2n−1)(2n+1)(2n+3))) S_n =C−(1/((2n+1)(2n+3)×2×2)) put n=1 S_1 =T_1 =(1/(15)) (1/(15))=C−(1/(4(3)(5))) C=(1/(15))+(1/(60))=((4+1)/(60))=(1/(12)) S_n =(1/(12))−(1/(4(2n+1)(2n+3))) checking the answer by follosing proof 1)S_1 =T_1 2)S_2 =T_1 +T_2 3)S_3 =T_1 +T_2 +T_3 T_n =(1/((2n−1)(2n+1)((2n+3))) T_1 =(1/(1×3×5))=(1/(15)) S_1 =(1/(12))−(1/(4×3×5))=(1/(15)) T_1 =S_1 proved T_2 =(1/(3×5×7))=(1/(105)) T_3 =(1/(5×7×9))=(1/(315)) T_1 +T_2 =(1/(15))+(1/(105))=(8/(105)) S_n =(1/(12))−(1/(4(2n+1)(2n+3))) S_2 =(1/(12))−(1/(4×5×7))=(8/(105)) so S_2 =T_1 +T_2 proved T_1 +T_2 +T_3 =(8/(105))+(1/(315))=(5/(63)) S_3 =(1/(12))−(1/(4×7×9))=(5/(63)) so my answer iscorrect...

$${T}_{{n}} =\frac{\mathrm{1}}{\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}\right\}\left\{\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{2}\right\}\left\{\mathrm{5}+\left({n}−\mathrm{1}\right)\mathrm{2}\right\}} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)×\mathrm{2}×\mathrm{2}} \\ $$$${put}\:{n}=\mathrm{1}\:\:\:{S}_{\mathrm{1}} ={T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{15}} \\ $$$$\frac{\mathrm{1}}{\mathrm{15}}={C}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{3}\right)\left(\mathrm{5}\right)} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{60}}=\frac{\mathrm{4}+\mathrm{1}}{\mathrm{60}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${checking}\:{the}\:{answer}\:\:{by}\:{follosing}\:{proof} \\ $$$$\left.\mathrm{1}\right){S}_{\mathrm{1}} ={T}_{\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){S}_{\mathrm{2}} ={T}_{\mathrm{1}} +{T}_{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){S}_{\mathrm{3}} ={T}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\left(\mathrm{2}{n}+\mathrm{3}\right)\right.} \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{15}}\:\:\:\:\:\:\:\:{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}×\mathrm{3}×\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{15}}\:\:{T}_{\mathrm{1}} ={S}_{\mathrm{1}} \:{proved} \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}×\mathrm{5}×\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{105}} \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{5}×\mathrm{7}×\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{315}} \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{105}}=\frac{\mathrm{8}}{\mathrm{105}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}×\mathrm{7}}=\frac{\mathrm{8}}{\mathrm{105}}\:\:{so}\:{S}_{\mathrm{2}} ={T}_{\mathrm{1}} +{T}_{\mathrm{2}} \:\:{proved} \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{105}}+\frac{\mathrm{1}}{\mathrm{315}}=\frac{\mathrm{5}}{\mathrm{63}} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}×\mathrm{7}×\mathrm{9}}=\frac{\mathrm{5}}{\mathrm{63}} \\ $$$${so}\:{my}\:{answer}\:{iscorrect}… \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18