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Question Number 9890 by konen last updated on 12/Jan/17

(1−(1/(16)))(1−(1/(25)))(1−(1/(36)))...(1−(1/(3600)))=?

$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{36}}\right)...\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3600}}\right)=? \\ $$

Answered by nume1114 last updated on 12/Jan/17

(1−(1/(16)))(1−(1/(25)))(1−(1/(36)))...(1−(1/(3600))) =(1−(1/4^2 ))(1−(1/5^2 ))(1−(1/6^2 ))...(1−(1/(60^2 ))) =((4^2 −1)/4^2 )×((5^2 −1)/5^2 )×((6^2 −1)/6^2 )×...×((60^2 −1)/(60^2 )) =((3×5)/4^2 )×((4×6)/5^2 )×((5×7)/6^2 )×((6×8)/7^2 )×...×((58×60)/(59^2 ))×((59×61)/(60^2 )) =((3×4×60×61)/(4^2 ×60^2 ))=((61)/(80))

$$\:\:\:\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{36}}\right)...\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3600}}\right) \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\right)...\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{60}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }×\frac{\mathrm{5}^{\mathrm{2}} −\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }×\frac{\mathrm{6}^{\mathrm{2}} −\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }×...×\frac{\mathrm{60}^{\mathrm{2}} −\mathrm{1}}{\mathrm{60}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}×\mathrm{5}}{\mathrm{4}^{\mathrm{2}} }×\frac{\mathrm{4}×\mathrm{6}}{\mathrm{5}^{\mathrm{2}} }×\frac{\mathrm{5}×\mathrm{7}}{\mathrm{6}^{\mathrm{2}} }×\frac{\mathrm{6}×\mathrm{8}}{\mathrm{7}^{\mathrm{2}} }×...×\frac{\mathrm{58}×\mathrm{60}}{\mathrm{59}^{\mathrm{2}} }×\frac{\mathrm{59}×\mathrm{61}}{\mathrm{60}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}×\mathrm{4}×\mathrm{60}×\mathrm{61}}{\mathrm{4}^{\mathrm{2}} ×\mathrm{60}^{\mathrm{2}} }=\frac{\mathrm{61}}{\mathrm{80}} \\ $$

Commented by RasheedSoomro last updated on 13/Jan/17

G^( O^(⌢) O^(⌢) ) D Technic !

$$\mathcal{G}^{\:\overset{\frown} {\mathcal{O}}\overset{\frown} {\mathcal{O}}} \mathcal{D}\:\:\mathcal{T}{echnic}\:! \\ $$

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