Question Number 150231 by Lekhraj last updated on 10/Aug/21
Answered by Ar Brandon last updated on 10/Aug/21
![f(x)=(1/(3^x +(√3))) ⇒f(1−x)=(1/(3^(1−x) +(√3))) =(3^x /(3+3^x (√3)))=(1/( (√3)))∙(3^x /(3^x +(√3))) ⇒f(x)+f(1−x)=((1+(3^x /( (√3))))/(3^x +(√3)))=(1/( (√3)))∙((3^x +(√3))/(3^x +(√3)))=(1/( (√3))) Σ_(x=1) ^6 [f(x)+f(1−x)]=(6/( (√3))) Ans:(√3)Σ_(x=1) ^6 [f(x)+f(1−x)]=6](Q150237.png)
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}−{x}} +\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}^{{x}} }{\mathrm{3}+\mathrm{3}^{{x}} \sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\mathrm{3}^{{x}} }{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}+\frac{\mathrm{3}^{{x}} }{\:\sqrt{\mathrm{3}}}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left[{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\right]=\frac{\mathrm{6}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Ans}:\sqrt{\mathrm{3}}\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left[{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\right]=\mathrm{6} \\ $$
Commented by Lekhraj last updated on 10/Aug/21
$$\mathrm{Very}\:\mathrm{nice}\:.\:\mathrm{Thank}\:\mathrm{you}. \\ $$
Commented by Ar Brandon last updated on 10/Aug/21
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{Sir} \\ $$