Question

Tính giá trị của biểu thức A=\(\dfrac{1}{2016^{-2016}+1}+\dfrac{1}{2016^{-2015}+1}+....+\dfrac{1}{2016^{-1}+1}+\dfrac{1}{2016^0+1}+\dfrac{1}{2016^1+1}+......+\dfrac{1}{2016^{2016}+1}\)

Answer 1

Lời giải:

Ta thấy:

\(\frac{1}{2016^x+1}+\frac{1}{2016^{-x}+1}=\frac{1}{2016^x+1}+\frac{1}{\frac{1}{2016^x}+1}=\frac{1}{2016^x+1}+\frac{2016^x}{1+2016^x}=\frac{2016^x+1}{2016^x+1}=1\)

Do đó:

\(A=\frac{1}{2016^{-2016}+1}+\frac{1}{2016^{-2015}+1}+...+\frac{1}{2016^{-1}+1}+\frac{1}{2016^0+1}+\frac{1}{2016^1+1}+...+\frac{1}{2016^{2016}+1}\)

\(=\underbrace{\left(\frac{1}{2016^{-2016}+1}+\frac{1}{2016^{2016}+1}\right)+\left(\frac{1}{2016^{-2015}+1}+\frac{1}{2016^{2015}+1}\right)+....+\left(\frac{1}{2016^{-1}+1}+\frac{1}{2016^{1}+1}\right)}_{ \text{2016 cặp}}+\frac{1}{2016^0+1}\)

\(=1.2016+\frac{1}{1+1}=2016+\frac{1}{2}=\frac{4033}{2}\)