Sum of function

Algebra Level 3

If f ( x ) = 2 2 + 4 x f(x)=\frac{2}{2+4^x} Find, A = f ( 1 101 ) + f ( 2 101 ) + f ( 3 101 ) + . . . + f ( 100 101 ) A=f\left(\frac{1}{101}\right)+f\left(\frac{2}{101}\right)+f\left(\frac{3}{101}\right)+...+f\left(\frac{100}{101}\right) .


The answer is 50.

Achmad Damanhuri by Achmad Damanhuri , 26, Indonesia

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1 solution

f ( x ) = 2 2 + 4 x = 1 1 + 2 2 x − 1 = 1 1 + 2 x + ( x − 1 ) It is interesting to note that, f ( 1 − x ) = 1 1 + 2 ( 1 − x ) − x = 1 1 + 2 1 − 2 x = 2 2 x − 1 1 + 2 2 x − 1 = 1 − 1 1 + 2 2 x − 1 = 1 − f ( x ) ⟹ f ( x ) + f ( 1 − x ) = 1 ( 1 ) A = ∑ n = 1 100 f ( n 101 ) = ∑ n = 1 100 f ( 101 − n 101 ) ⟹ 2 A = ∑ n = 1 100 f ( n 101 ) + f ( 101 − n 101 ) = ∑ n = 1 100 1 From ( 1 ) ⟹ A = 50 \begin{aligned}\\ f(x)&=\dfrac{2}{2+4^x}\\ &=\dfrac{1}{1+2^{2x-1}}\\ &=\dfrac{1}{1+2^{x+(x-1)}}\\ \text{It is interesting to note that,}\\ f(1-x)&=\dfrac{1}{1+2^{(1-x)-x}}\\ &=\dfrac{1}{1+2^{1-2x}}\\ &=\dfrac{2^{2x-1}}{1+2^{2x-1}}\\ &=1-\dfrac{1}{1+2^{2x-1}}\\ &=1-f(x)\\\\ \implies f(x)&+f(1-x)=1 \hspace{4mm}\color{#3D99F6}\small(1) \\ \text{A}&=\sum_{n=1}^{100} f\left(\dfrac{n}{101}\right)\\ &=\sum_{n=1}^{100} f\left(\dfrac{101-n}{101}\right)\\ \implies 2\text{A}&=\sum_{n=1}^{100} f\left(\dfrac{n}{101}\right)+f\left(\dfrac{101-n}{101}\right)\\ &=\sum_{n=1}^{100} \hspace{2mm}1 \hspace{8mm}\color{#3D99F6}\small\text{From }(1)\\ \implies\text{A}&=\color{#EC7300}\boxed{\color{#333333}50}\end{aligned}

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