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Question Number 44986 by Tawa1 last updated on 07/Oct/18

Find the sum of the nth term of the series: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

pls write the answer of questiins snd source of question...

$${pls}\:{write}\:{the}\:{answer}\:{of}\:{questiins}\:{snd}\:{source}\:{of}\: \\ $$$${question}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

T_n =(1/({1+(n−1)×3}(2+(n−1)×3}{3+(n−1)×3})) T_n =(1/((3n−2)(3n−1)(3n))) here i am using formula tl be attached later S_n =C−(1/((3n−1)3n×1×2)) (1/(1×2×3))=C−(1/(2×3×2)) C=(1/6)+(1/(12))=((2+1)/(12))=(1/4) S_n =(1/4)−(1/(2(3n−1)(3n))) Pls check recheck... S_2 =(1/4)−(1/(2×5×6))=((15−1)/(60))=(7/(30)) but (1/(1×2×3))+(1/(4×5×6))=(1/6)+(1/(120))=((20+1)/(120))=(7/(40)) n=3 S_3 =(1/4)−(1/(2(8)(9)))=((36−1)/(144))=((35)/(144)) (1/6)+(1/(120))+(1/(210)) some problem is there i can not detect...

$${T}_{{n}} =\frac{\mathrm{1}}{\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)×\mathrm{3}\right\}\left(\mathrm{2}+\left({n}−\mathrm{1}\right)×\mathrm{3}\right\}\left\{\mathrm{3}+\left({n}−\mathrm{1}\right)×\mathrm{3}\right\}} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{2}\right)\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)} \\ $$$${here}\:{i}\:{am}\:{using}\:{formula}\:{tl}\:{be}\:{attached}\:{later} \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\mathrm{3}{n}×\mathrm{1}×\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}={C}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{2}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)} \\ $$$${Pls}\:{check}\: \\ $$$${recheck}... \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{5}×\mathrm{6}}=\frac{\mathrm{15}−\mathrm{1}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{30}} \\ $$$${but}\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}×\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{120}}=\frac{\mathrm{20}+\mathrm{1}}{\mathrm{120}}=\frac{\mathrm{7}}{\mathrm{40}} \\ $$$${n}=\mathrm{3} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{8}\right)\left(\mathrm{9}\right)}=\frac{\mathrm{36}−\mathrm{1}}{\mathrm{144}}=\frac{\mathrm{35}}{\mathrm{144}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{210}}\:\:\:\:{some}\:{problem}\:{is}\:{there}\:{i}\:{can}\:{not}\:{detect}... \\ $$

Commented by Tawa1 last updated on 07/Oct/18

No answer there sir. Thanks for everytime sir. God bless you.

$$\mathrm{No}\:\mathrm{answer}\:\mathrm{there}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{everytime}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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