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Show that,`(1+5^(-1))(1+5^(-2))(1+5^(-4))(1+5^(-8))"....."(1+5^(-2n))=(5)/(4)(1-5^(-2(n+1)))`. |
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Answer» `LHS=(1+5^(-1))(1+5^(-2))(1+5^(-4))(1+5^(-8))"....."(1+5^(-2n))` `=(1+(1)/(5))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))` `=((1-(1)/(5)))/((1-(1)/(5)))(1+(1)/(5))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))` `=(5)/(4)[(1+(1)/(5^(2)))(1+(1)/(5^(2)))(1+(1)/(5^(4)))"......."(1+(1)/(5^(2n)))]` `" " vdots " " vdots " "vdots" "` `=(5)/(4)(1-(1)/(5^(2^(n+1))))=(5)/(4)(1-5^(2^(n+1)))=RHS`. |
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