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Given-that-N-625-1-9-5-2-1-9-8-2-1-9-11-2-1-9-125-2-Find-the-sum-of-the-digits-of-N-

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Question Number 114570 by ZiYangLee last updated on 19/Sep/20
Given that N=625(1−(9/5^2 ))(1−(9/8^2 ))(1−(9/(11^2 )))...(1−(9/(125^2 ))) Find the sum of the digits of N.
$$\mathrm{Given}\:\mathrm{that}\: \\ $$$${N}=\mathrm{625}\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{5}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{8}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{11}^{\mathrm{2}} }\right)…\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{125}^{\mathrm{2}} }\right) \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:{N}. \\ $$
Answered by floor(10²Eta[1]) last updated on 19/Sep/20
N=625(((2.8)/5^2 ))(((5.11)/8^2 ))(((8.14)/(11^2 )))(((11.17)/(14^2 )))...(((122.128)/(125^2 ))) N=625((2/5))(((128)/(125)))=2.128=256 2+5+6=13
$$\mathrm{N}=\mathrm{625}\left(\frac{\mathrm{2}.\mathrm{8}}{\mathrm{5}^{\mathrm{2}} }\right)\left(\frac{\mathrm{5}.\mathrm{11}}{\mathrm{8}^{\mathrm{2}} }\right)\left(\frac{\mathrm{8}.\mathrm{14}}{\mathrm{11}^{\mathrm{2}} }\right)\left(\frac{\mathrm{11}.\mathrm{17}}{\mathrm{14}^{\mathrm{2}} }\right)…\left(\frac{\mathrm{122}.\mathrm{128}}{\mathrm{125}^{\mathrm{2}} }\right) \\ $$$$\mathrm{N}=\mathrm{625}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\left(\frac{\mathrm{128}}{\mathrm{125}}\right)=\mathrm{2}.\mathrm{128}=\mathrm{256} \\ $$$$\mathrm{2}+\mathrm{5}+\mathrm{6}=\mathrm{13} \\ $$
Answered by Olaf last updated on 19/Sep/20
N = 625Π_(n=1) ^(41) (1−(9/((3n+2)^2 ))) N = 625Π_(n=1) ^(41) ((((3n+2)^2 −9)/((3n+2)^2 ))) N = 625Π_(n=1) ^(41) (((9n^2 +12n−5)/((3n+2)^2 ))) 9n^2 +12n−5 = 0 Δ′ = 36−9(−5) = 36+45 = 81 1st root : ((−6−9)/9) = −(5/3) 2nd root : ((−6+9)/9) = (1/3) N = 625Π_(n=1) ^(41) ((9(n+(5/3))(n−(1/3)))/((3n+2)^2 )) N = 625Π_(n=1) ^(41) (((3n+5)(3n−1))/((3n+2)^2 )) N = 625Π_(n=1) ^(41) (((3(n+1)+2)(3(n−1)+2))/((3n+2)^2 )) u_n = 3n+2, n≥0 N = 625((Π_(n=1) ^(41) u_(n+1) ×Π_(n=1) ^(41) u_(n−1) )/((Π_(n=1) ^(41) u_n )^2 )) N = 625((Π_(n=2) ^(42) u_n ×Π_(n=0) ^(40) u_n )/((Π_(n=1) ^(41) u_n )^2 )) N = 625×(u_(42) /u_1 )×(u_0 /u_(41) ) u_0 = 3(0)+2 = 2 u_1 = 3(1)+2 = 5 u_(41) = 3(41)+2 = 125 u_(42) = 3(42)+2 = 128 N = 625×((128)/5)×(2/(125)) N = 256 sum of the digits of N is : 2 + 5 + 6 = 13
$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{9}}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\left(\frac{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9}}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\left(\frac{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{12}{n}−\mathrm{5}}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{9}{n}^{\mathrm{2}} +\mathrm{12}{n}−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\Delta'\:=\:\mathrm{36}−\mathrm{9}\left(−\mathrm{5}\right)\:=\:\mathrm{36}+\mathrm{45}\:=\:\mathrm{81} \\ $$$$\mathrm{1st}\:\mathrm{root}\::\:\frac{−\mathrm{6}−\mathrm{9}}{\mathrm{9}}\:=\:−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{2nd}\:\mathrm{root}\::\:\frac{−\mathrm{6}+\mathrm{9}}{\mathrm{9}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\frac{\mathrm{9}\left({n}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left({n}−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\frac{\left(\mathrm{3}{n}+\mathrm{5}\right)\left(\mathrm{3}{n}−\mathrm{1}\right)}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{N}\:=\:\mathrm{625}\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}\frac{\left(\mathrm{3}\left({n}+\mathrm{1}\right)+\mathrm{2}\right)\left(\mathrm{3}\left({n}−\mathrm{1}\right)+\mathrm{2}\right)}{\left(\mathrm{3}{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${u}_{{n}} \:=\:\mathrm{3}{n}+\mathrm{2},\:{n}\geqslant\mathrm{0} \\ $$$$\mathrm{N}\:=\:\mathrm{625}\frac{\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}{u}_{{n}+\mathrm{1}} ×\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}{u}_{{n}−\mathrm{1}} }{\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}{u}_{{n}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{N}\:=\:\mathrm{625}\frac{\underset{{n}=\mathrm{2}} {\overset{\mathrm{42}} {\prod}}{u}_{{n}} ×\underset{{n}=\mathrm{0}} {\overset{\mathrm{40}} {\prod}}{u}_{{n}} }{\left(\underset{{n}=\mathrm{1}} {\overset{\mathrm{41}} {\prod}}{u}_{{n}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{N}\:=\:\mathrm{625}×\frac{{u}_{\mathrm{42}} }{{u}_{\mathrm{1}} }×\frac{{u}_{\mathrm{0}} }{{u}_{\mathrm{41}} } \\ $$$${u}_{\mathrm{0}} \:=\:\mathrm{3}\left(\mathrm{0}\right)+\mathrm{2}\:=\:\mathrm{2} \\ $$$${u}_{\mathrm{1}} \:=\:\mathrm{3}\left(\mathrm{1}\right)+\mathrm{2}\:=\:\mathrm{5} \\ $$$${u}_{\mathrm{41}} \:=\:\mathrm{3}\left(\mathrm{41}\right)+\mathrm{2}\:=\:\mathrm{125} \\ $$$${u}_{\mathrm{42}} \:=\:\mathrm{3}\left(\mathrm{42}\right)+\mathrm{2}\:=\:\mathrm{128} \\ $$$$\mathrm{N}\:=\:\mathrm{625}×\frac{\mathrm{128}}{\mathrm{5}}×\frac{\mathrm{2}}{\mathrm{125}} \\ $$$$\mathrm{N}\:=\:\mathrm{256} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{N}\:\mathrm{is}\:: \\ $$$$\mathrm{2}\:+\:\mathrm{5}\:+\:\mathrm{6}\:=\:\mathrm{13} \\ $$$$ \\ $$$$ \\ $$

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