Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 45316 by Tawa1 last updated on 11/Oct/18

I heard this sum can be close using the Digamma function. Please help me use it. i don′t know it. sum of nth term: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...

$$\mathrm{I}\:\mathrm{heard}\:\mathrm{this}\:\mathrm{sum}\:\mathrm{can}\:\mathrm{be}\:\mathrm{close}\:\mathrm{using}\:\mathrm{the}\:\mathrm{Digamma}\:\mathrm{function}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{use}\:\mathrm{it}.\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{it}.\:\:\: \\ $$$$\:\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{nth}\:\mathrm{term}:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:...\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/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 some formula mentioned Hall and knight Algebra... S_n =C−(1/((3n−1)(3n)×2×1)) n=1 (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))) recheck... S_1 =(1/4)−(1/(2×2×3))=((3−1)/(12))=(1/6)=T_1 =(1/(1×2×3))=(1/6) conform S_2 =(1/4)−(1/(2×5×6))=((15−1)/(60))=((14)/(60))=(7/(30)) T_2 =S_2 −S_1 =(7/(30))−(1/6)=((7−5)/(30))=(1/(15)) but T_2 =(1/(120)) So my answer is not correct... pls menhion the source of this problem S_n =(1/(1×2×3))+(1/(4×5×6))+(1/(7×8×9))+... =

$${T}_{{n}} =\frac{\mathrm{1}}{\left\{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{2}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}\left\{\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{3}\right\}} \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{2}\right)\left(\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)\right.} \\ $$$${here}\:{i}\:{am}\:{using}\:{some}\:{formula}\:{mentioned} \\ $$$${Hall}\:{and}\:{knight}\:{Algebra}... \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)×\mathrm{2}×\mathrm{1}} \\ $$$${n}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}={C}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{2}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}\right)} \\ $$$${recheck}... \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}×\mathrm{3}}=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{6}}={T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${conform} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{5}×\mathrm{6}}=\frac{\mathrm{15}−\mathrm{1}}{\mathrm{60}}=\frac{\mathrm{14}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{30}} \\ $$$${T}_{\mathrm{2}} ={S}_{\mathrm{2}} −{S}_{\mathrm{1}} =\frac{\mathrm{7}}{\mathrm{30}}−\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{7}−\mathrm{5}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{15}} \\ $$$${but}\:{T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{120}} \\ $$$${So}\:{my}\:{answer}\:{is}\:{not}\:{correct}... \\ $$$${pls}\:{menhion}\:{the}\:{source}\:{of}\:{this}\:{problem} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}×\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}×\mathrm{8}×\mathrm{9}}+... \\ $$$$= \\ $$

Commented by Tawa1 last updated on 12/Oct/18

The question was asked online. But sir, when can i apply the formular you used to find the sum.

$$\mathrm{The}\:\mathrm{question}\:\mathrm{was}\:\mathrm{asked}\:\mathrm{online}.\:\: \\ $$$$\mathrm{But}\:\mathrm{sir},\:\mathrm{when}\:\mathrm{can}\:\mathrm{i}\:\mathrm{apply}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{you}\:\mathrm{used}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

ok.. i am posting...

$${ok}..\:{i}\:{am}\:{posting}... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

Commented by Tawa1 last updated on 12/Oct/18

Wow, God bless you sir. I really appreciate your effort.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com