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B-3-4-3-2-1-3-7-3-4-4-4-2-1-4-7-4-10-4-10-2-1-10-7-1-Find-B-1-220-

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Question Number 216178 by hardmath last updated on 29/Jan/25
B = ((3^4 + 3^2 + 1)/(3^7 - 3)) + ((4^4 + 4^2 + 1)/(4^7 - 4)) + ... + ((10^4 + 10^2 + 1)/(10^7 - 1)) Find: B + (1/(220)) = ?
$$\mathrm{B}\:=\:\frac{\mathrm{3}^{\mathrm{4}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{3}^{\mathrm{7}} \:-\:\mathrm{3}}\:+\:\frac{\mathrm{4}^{\mathrm{4}} \:+\:\mathrm{4}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{4}^{\mathrm{7}} \:-\:\mathrm{4}}\:+\:…\:+\:\frac{\mathrm{10}^{\mathrm{4}} \:+\:\mathrm{10}^{\mathrm{2}} \:+\:\mathrm{1}}{\mathrm{10}^{\mathrm{7}} \:-\:\mathrm{1}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{B}\:+\:\frac{\mathrm{1}}{\mathrm{220}}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jan/25
B=Σ_(x=3) ^(10) ((x^4 +x^2 +1)/(x^7 −x)) ; x∈N =Σ_(x=3) ^(10) (((x^2 +x+1)(x^2 −x+1))/(x(x^6 −1))) =Σ_(x=3) ^(10) (((x^2 +x+1)(x^2 −x+1))/( x(x−1)(x+1)(x^2 +x+1)(x^2 −x+1))) =Σ_(x=3) ^(10) (1/( x(x−1)(x+1))) =Σ_(x=3) ^(10) (1/( x(x^2 −1))) =(1/(3(3^2 −1)))+(1/(4(4^2 −1)))+(1/(5(5^2 −1)))+(1/(6(6^2 −1)))+(1/(7(7^2 −1)))+(1/(8(8^2 −1)))+(1/(9(9^2 −1)))+(1/(10(10^2 −1))) =(1/(24))+(1/(60))+(1/(120))+(1/(210))+(1/(336))+(1/(504))+(1/(720))+(1/(990)) =((13)/(165)) B+(1/(220))=((13)/(165))+(1/(220))=(1/(12))
$${B}=\underset{{x}=\mathrm{3}} {\overset{\mathrm{10}} {\Sigma}}\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{7}} −{x}}\:;\:{x}\in\mathbb{N} \\ $$$$\:=\underset{{x}=\mathrm{3}} {\overset{\mathrm{10}} {\Sigma}}\frac{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{{x}\left({x}^{\mathrm{6}} −\mathrm{1}\right)} \\ $$$$\:=\underset{{x}=\mathrm{3}} {\overset{\mathrm{10}} {\Sigma}}\:\frac{\cancel{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}}{\:\:{x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\cancel{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}} \\ $$$$\:=\underset{{x}=\mathrm{3}} {\overset{\mathrm{10}} {\Sigma}}\:\frac{\mathrm{1}}{\:\:{x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\ $$$$\:=\underset{{x}=\mathrm{3}} {\overset{\mathrm{10}} {\Sigma}}\:\frac{\mathrm{1}}{\:\:{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{5}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{6}\left(\mathrm{6}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{7}\left(\mathrm{7}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{8}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{9}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{10}\left(\mathrm{10}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{60}}+\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{210}}+\frac{\mathrm{1}}{\mathrm{336}}+\frac{\mathrm{1}}{\mathrm{504}}+\frac{\mathrm{1}}{\mathrm{720}}+\frac{\mathrm{1}}{\mathrm{990}} \\ $$$$=\frac{\mathrm{13}}{\mathrm{165}} \\ $$$${B}+\frac{\mathrm{1}}{\mathrm{220}}=\frac{\mathrm{13}}{\mathrm{165}}+\frac{\mathrm{1}}{\mathrm{220}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Answered by mr W last updated on 29/Jan/25
a_n =((n^4 +n^2 +1)/(n^7 −n)) =((n^4 +2n^2 +1−n^2 )/(n(n^6 −1))) =(((n^2 +1)^2 −n^2 )/(n(n^3 +1)(n^3 −1))) =(((n^2 −n+1)(n^2 +n+1))/(n(n+1)(n^2 −n+1)(n−1)(n^2 +n+1))) =(1/(n(n+1)(n−1))) =(1/2)×((n+1−(n−1))/((n−1)n(n+1))) =(1/2)[(1/((n−1)n))−(1/(n(n+1)))] B=a_3 +a_4 +...+a_(10) =(1/2)[((1/(2×3))−(1/(3×4)))+((1/(3×4))−(1/(4×5)))+...+((1/(9×10))−(1/(10×11)))] =(1/2)((1/(2×3))−(1/(10×11))) =(1/(12))−(1/(220)) ⇒B+(1/(220))=(1/(12)) ✓
$${a}_{{n}} =\frac{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}{{n}^{\mathrm{7}} −{n}} \\ $$$$\:\:\:\:=\frac{{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}−{n}^{\mathrm{2}} }{{n}\left({n}^{\mathrm{6}} −\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\left({n}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} }{{n}\left({n}^{\mathrm{3}} +\mathrm{1}\right)\left({n}^{\mathrm{3}} −\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)}{{n}\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{n}+\mathrm{1}−\left({n}−\mathrm{1}\right)}{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}}−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right] \\ $$$$ \\ $$$${B}={a}_{\mathrm{3}} +{a}_{\mathrm{4}} +…+{a}_{\mathrm{10}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{9}×\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{10}×\mathrm{11}}\right)\right] \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{10}×\mathrm{11}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{220}} \\ $$$$ \\ $$$$\Rightarrow{B}+\frac{\mathrm{1}}{\mathrm{220}}=\frac{\mathrm{1}}{\mathrm{12}}\:\checkmark \\ $$
Commented by hardmath last updated on 29/Jan/25
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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