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Question Number 206066 by hardmath last updated on 06/Apr/24

Find: (3/4) ∙ (8/9) ∙ ((15)/(16)) ∙ ... ∙ ((120)/(121)) = ?

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{8}}{\mathrm{9}}\:\centerdot\:\frac{\mathrm{15}}{\mathrm{16}}\:\centerdot\:...\:\centerdot\:\frac{\mathrm{120}}{\mathrm{121}}\:=\:? \\ $$

Answered by mr W last updated on 06/Apr/24

Π_(n=2) ^(11) ((n^2 −1)/n^2 ) =Π_(n=2) ^(11) (((n−1)/n)×((n+1)/n)) =Π_(n=2) ^(11) ((n−1)/n)×Π_(n=2) ^(11) ((n+1)/n) =((1/2)×(2/3)×(3/4)...×((10)/(11)))×((3/2)×(4/3)×(5/4)×...×((12)/(11))) =(1/(11))×((12)/2) =(6/(11))

$$\underset{{n}=\mathrm{2}} {\overset{\mathrm{11}} {\prod}}\frac{{n}^{\mathrm{2}} −\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\mathrm{11}} {\prod}}\left(\frac{{n}−\mathrm{1}}{{n}}×\frac{{n}+\mathrm{1}}{{n}}\right) \\ $$$$=\underset{{n}=\mathrm{2}} {\overset{\mathrm{11}} {\prod}}\frac{{n}−\mathrm{1}}{{n}}×\underset{{n}=\mathrm{2}} {\overset{\mathrm{11}} {\prod}}\frac{{n}+\mathrm{1}}{{n}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}...×\frac{\mathrm{10}}{\mathrm{11}}\right)×\left(\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{4}}×...×\frac{\mathrm{12}}{\mathrm{11}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{11}}×\frac{\mathrm{12}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}}{\mathrm{11}} \\ $$

Commented by hardmath last updated on 06/Apr/24

cool dear professor thankyou

$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$

Answered by MATHEMATICSAM last updated on 06/Apr/24

We can see that each term in terms of n is ((n^2 − 1)/n^2 ) where n = 2, 3,4, ... , 11 respectively. ((n^2 − 1)/n^2 ) = (((n + 1)(n − 1))/(n.n)) (3/4) × (8/9) × ((15)/(16)) × ... × ((120)/(121)) = ((3 × 1)/(2 × 2)) × ((4 × 2)/(3 × 3)) × ((5 × 3)/(4 × 4)) × ... × ((12 × 10)/(11 × 11)) = (1/2) × ((12)/(11)) = (6/(11)) (Ans)

$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that}\:\mathrm{each}\:\mathrm{term}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\ $$$${n}\:\mathrm{is}\:\frac{{n}^{\mathrm{2}} \:−\:\mathrm{1}}{{n}^{\mathrm{2}} }\:\mathrm{where}\:{n}\:=\:\mathrm{2},\:\mathrm{3},\mathrm{4},\:...\:,\:\mathrm{11}\: \\ $$$$\mathrm{respectively}. \\ $$$$\frac{{n}^{\mathrm{2}} \:−\:\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\frac{\left({n}\:+\:\mathrm{1}\right)\left({n}\:−\:\mathrm{1}\right)}{{n}.{n}}\: \\ $$$$ \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:×\:\frac{\mathrm{8}}{\mathrm{9}}\:×\:\frac{\mathrm{15}}{\mathrm{16}}\:×\:...\:×\:\frac{\mathrm{120}}{\mathrm{121}} \\ $$$$=\:\frac{\cancel{\mathrm{3}}\:×\:\mathrm{1}}{\mathrm{2}\:×\:\cancel{\mathrm{2}}}\:×\:\frac{\cancel{\mathrm{4}}\:×\:\cancel{\mathrm{2}}}{\cancel{\mathrm{3}}\:×\:\cancel{\mathrm{3}}}\:×\:\frac{\cancel{\mathrm{5}}\:×\:\cancel{\mathrm{3}}}{\cancel{\mathrm{4}}\:×\:\cancel{\mathrm{4}}}\:×\:...\:×\:\frac{\mathrm{12}\:×\:\cancel{\mathrm{10}}}{\cancel{\mathrm{11}}\:×\:\mathrm{11}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\mathrm{12}}{\mathrm{11}}\:=\:\frac{\mathrm{6}}{\mathrm{11}}\:\left(\mathrm{Ans}\right)\:\: \\ $$

Commented by hardmath last updated on 06/Apr/24

thank you dear professor cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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