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Question Number 209357 by mnjuly1970 last updated on 08/Jul/24
Evaluate : B_n = Π_(k=3) ^n (( k^( 2) −1)/(k^2 + k −6))= ?
$$ \\ $$$$\:\:\:\:\:\:{Evaluate}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{B}_{{n}} =\:\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\:\frac{\:{k}^{\:\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} \:+\:{k}\:−\mathrm{6}}=\:? \\ $$
Answered by Spillover last updated on 08/Jul/24
B_n = Π_(k=3) ^n (( (k−1)(k+1))/((k−2)(k+3))) B_n =(((3−1)(3+1))/((3−2)(3+3))).(((4−1)(4+1))/((4−2)(4+3))).(((5−1)(5+1))/((5−2)(5+3)))......(((n−1)(n+1))/((n−2)(n+3))). B_n =((2.4)/(1.6)).((3.5)/(2.7)).((4.6)/(3.8))......(((n−1)(n+1))/((n−2)(n+3))) B_n =((2.4.3.5.4.6.....(n−1)(n+1))/(1.6.2.7.3.8.....(n−2)(n+3))) B_n =((2.(n+1))/(1.(n+3))) B_n =((2(n+1))/((n+3)))
$$\:\:\:\:\:\mathrm{B}_{{n}} =\:\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\:\frac{\:\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{\left({k}−\mathrm{2}\right)\left({k}+\mathrm{3}\right)} \\ $$$${B}_{{n}} =\frac{\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}+\mathrm{1}\right)}{\left(\mathrm{3}−\mathrm{2}\right)\left(\mathrm{3}+\mathrm{3}\right)}.\frac{\left(\mathrm{4}−\mathrm{1}\right)\left(\mathrm{4}+\mathrm{1}\right)}{\left(\mathrm{4}−\mathrm{2}\right)\left(\mathrm{4}+\mathrm{3}\right)}.\frac{\left(\mathrm{5}−\mathrm{1}\right)\left(\mathrm{5}+\mathrm{1}\right)}{\left(\mathrm{5}−\mathrm{2}\right)\left(\mathrm{5}+\mathrm{3}\right)}……\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\left({n}−\mathrm{2}\right)\left({n}+\mathrm{3}\right)}. \\ $$$${B}_{{n}} =\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{6}}.\frac{\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{7}}.\frac{\mathrm{4}.\mathrm{6}}{\mathrm{3}.\mathrm{8}}……\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\left({n}−\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$${B}_{{n}} =\frac{\mathrm{2}.\mathrm{4}.\mathrm{3}.\mathrm{5}.\mathrm{4}.\mathrm{6}…..\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\mathrm{1}.\mathrm{6}.\mathrm{2}.\mathrm{7}.\mathrm{3}.\mathrm{8}…..\left({n}−\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$${B}_{{n}} =\frac{\mathrm{2}.\left({n}+\mathrm{1}\right)}{\mathrm{1}.\left({n}+\mathrm{3}\right)} \\ $$$${B}_{{n}} =\frac{\mathrm{2}\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{3}\right)} \\ $$$$ \\ $$
Commented by MM42 last updated on 08/Jul/24
for n=5 B_n =((2×4)/(1×6))×((3×5)/(2×7))×((4×6)/(3×8))=((10)/7)≠((12)/8)
$${for}\:{n}=\mathrm{5} \\ $$$${B}_{{n}} =\frac{\mathrm{2}×\mathrm{4}}{\mathrm{1}×\mathrm{6}}×\frac{\mathrm{3}×\mathrm{5}}{\mathrm{2}×\mathrm{7}}×\frac{\mathrm{4}×\mathrm{6}}{\mathrm{3}×\mathrm{8}}=\frac{\mathrm{10}}{\mathrm{7}}\neq\frac{\mathrm{12}}{\mathrm{8}} \\ $$$$ \\ $$
Answered by MM42 last updated on 08/Jul/24
((k^2 −1)/(k^2 +k−6))=(((k−1)(k+1))/((k−2)(k+3))) B_n =(((2×4)/(1×6)))(((3×5)/(2×7)))(((4×6)/(3×8)))...((((n−3)(n−1))/((n−4)(n+1))))((((n−2)(n))/((n−3)(n+2))))((((n−1)(n+1))/((n−2)(n+3)))) =((4×5(n−1))/(n(n+3))) ✓
$$\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} +{k}−\mathrm{6}}=\frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{\left({k}−\mathrm{2}\right)\left({k}+\mathrm{3}\right)} \\ $$$${B}_{{n}} =\left(\frac{\mathrm{2}×\mathrm{4}}{\mathrm{1}×\mathrm{6}}\right)\left(\frac{\mathrm{3}×\mathrm{5}}{\mathrm{2}×\mathrm{7}}\right)\left(\frac{\mathrm{4}×\mathrm{6}}{\mathrm{3}×\mathrm{8}}\right)…\left(\frac{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{1}\right)}{\left({n}−\mathrm{4}\right)\left({n}+\mathrm{1}\right)}\right)\left(\frac{\left({n}−\mathrm{2}\right)\left({n}\right)}{\left({n}−\mathrm{3}\right)\left({n}+\mathrm{2}\right)}\right)\left(\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\left({n}−\mathrm{2}\right)\left({n}+\mathrm{3}\right)}\right) \\ $$$$=\frac{\mathrm{4}×\mathrm{5}\left({n}−\mathrm{1}\right)}{{n}\left({n}+\mathrm{3}\right)}\:\checkmark \\ $$
Answered by 0670322918 last updated on 09/Jul/24
Evaluate : B_n = Π_(k=3) ^n (( k^( 2) −1)/(k^2 + k −6))= ? B_n =Π_(k=3) ^n (((k−1)(k+1))/((k−2)(k+3)))=((n−1)/1)×Π_(k=3) ^n (((k+1)/(k+2))×((k+2)/(k+3))) B_n =(n−1)×(4/(n+2))×(5/(n+3))=((20(n−1))/((n+2)(n+3)))
$$ \\ $$$$\:\:\:\:\:\:{Evaluate}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{B}_{{n}} =\:\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\:\frac{\:{k}^{\:\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} \:+\:{k}\:−\mathrm{6}}=\:? \\ $$$${B}_{{n}} =\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{\left({k}−\mathrm{2}\right)\left({k}+\mathrm{3}\right)}=\frac{{n}−\mathrm{1}}{\mathrm{1}}×\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\left(\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}}×\frac{{k}+\mathrm{2}}{{k}+\mathrm{3}}\right) \\ $$$${B}_{{n}} =\left({n}−\mathrm{1}\right)×\frac{\mathrm{4}}{{n}+\mathrm{2}}×\frac{\mathrm{5}}{{n}+\mathrm{3}}=\frac{\mathrm{20}\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$

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