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Question Number 114570 by ZiYangLee last updated on 19/Sep/20

$$\mathrm{Given}\mathrm{that}$$$$N=625(1-\frac{9}{5^2})(1-\frac{9}{8^2})(1-\frac{9}{{11}^2})...(1-\frac{9}{{125}^2})$$$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathrm{of}N.$$

Answered by floor(10²Eta[1]) last updated on 19/Sep/20

$$\mathrm{N}=625\left(\frac{2.8}{5^2}\right)\left(\frac{5.11}{8^2}\right)\left(\frac{8.14}{{11}^2}\right)\left(\frac{11.17}{{14}^2}\right)...\left(\frac{122.128}{{125}^2}\right)$$$$\mathrm{N}=625\left(\frac{2}{5}\right)\left(\frac{128}{125}\right)=2.128=256$$$$2+5+6=13$$

Answered by Olaf last updated on 19/Sep/20

$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}(1-\frac{9}{{(3n+2)}^2})$$$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}\left(\frac{{(3n+2)}^2-9}{{(3n+2)}^2}\right)$$$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}\left(\frac{9n^2+12n-5}{{(3n+2)}^2}\right)$$$$9n^2+12n-5=0$$$${\mathrm{\Delta }}^{\prime }=36-9(-5)=36+45=81$$$$1\mathrm{st}\mathrm{root}:\frac{-6-9}{9}=-\frac{5}{3}$$$$2\mathrm{nd}\mathrm{root}:\frac{-6+9}{9}=\frac{1}{3}$$$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}\frac{9(n+\frac{5}{3})(n-\frac{1}{3})}{{(3n+2)}^2}$$$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}\frac{(3n+5)(3n-1)}{{(3n+2)}^2}$$$$\mathrm{N}=625\underset{n=1}{\prod ^{41}}\frac{(3(n+1)+2)(3(n-1)+2)}{{(3n+2)}^2}$$$$u_n=3n+2,n⩾0$$$$\mathrm{N}=625\frac{\underset{n=1}{\prod ^{41}}{u}_{n+1}\times \underset{n=1}{\prod ^{41}}{u}_{n-1}}{{\left(\underset{n=1}{\prod ^{41}}{u}_n\right)}^2}$$$$\mathrm{N}=625\frac{\underset{n=2}{\prod ^{42}}{u}_n\times \underset{n=0}{\prod ^{40}}{u}_n}{{\left(\underset{n=1}{\prod ^{41}}{u}_n\right)}^2}$$$$\mathrm{N}=625\times \frac{u_{42}}{u_1}\times \frac{u_0}{u_{41}}$$$$u_0=3\left(0\right)+2=2$$$$u_1=3\left(1\right)+2=5$$$$u_{41}=3\left(41\right)+2=125$$$$u_{42}=3\left(42\right)+2=128$$$$\mathrm{N}=625\times \frac{128}{5}\times \frac{2}{125}$$$$\mathrm{N}=256$$$$\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathrm{of}\mathrm{N}\mathrm{is}:$$$$2+5+6=13$$$$$$$$$$

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