Question

Cho \(f\left(x\right)=x^2+x\)

Tính \(\dfrac{1}{f\left(1\right)}+\dfrac{1}{f\left(2\right)}+\dfrac{1}{f\left(3\right)}+...+\dfrac{1}{f\left(2014\right)}+\dfrac{1}{f\left(2015\right)}\)

Answer 1

Lời giải:

Ta có:

\(f(x)=x^2+x\Rightarrow \frac{1}{f(x)}=\frac{1}{x^2+x}=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\)

Do đó:

\(\frac{1}{f(1)}=1-\frac{1}{2}\)

\(\frac{1}{f(2)}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{f(3)}=\frac{1}{3}-\frac{1}{4}\)

......

\(\frac{1}{f(2014)}=\frac{1}{2014}-\frac{1}{2015}\)

\(\frac{1}{f(2015)}=\frac{1}{2015}-\frac{1}{2016}\)

Cộng theo vế:
\(\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(2014)}+\frac{1}{f(2015)}=1-\frac{1}{2016}\)

\(=\frac{2015}{2016}\)