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If `f(x)=(a^x)/(a^x+sqrt(a ,)),(a >0),`then find the value of`sum_(r=1)^(2n1)2f(r/(2n))` |
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Answer» `f(x)=(a^(x))/(a^(x)+sqrt(a))` or `f(1-x)=(a^(1-x))/(a^(1-x)+sqrt(a))` `=(a^(1))/(a^(1)+sqrt(a)a^(x))` `=(sqrt(a))/(sqrt(a)+a^(x))` ` or f(x)+f(1-x)=1` ` :. f((1)/(2))=(1)/(2)` so, `sum_(r=1)^(2n-1)2f((r)/(2n))=2[f((1)/(2n))+f((2)/(2n))+ ... +f((n-1)/(2n))+f((n)/(2n))+f((n+1)/(2n))+ ... +f((2n-1)/(2n))]` `=2{[f((1)/(2n))+f((2n-1)/(2n))]+[f((2)/(2n))+f((2n-2)/(2n))]+ ... +[f((n-1)/(2n))+f((n+1)/(2n))]+f((1)/(2))}` `=2{[f((1)/(2n))+f(1-(1)/(2n))]+[f((2)/(2n))+f(1-(2)/(2n))]+ ... +[f((n-1)/(2n))+f(1-(n-1)/(2n))]+(1)/(2)}` `=2[1+1+1+...+(n-1)" times "]+1` `=2n-1` |
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