Tính \(f\left(1\right)\)
\(f\left(x\right)=1+x^3+x^5+x^7+...+x^{101}\)
\(\Rightarrow f\left(1\right)=1+1^3+1^5+1^7+...+1^{101}\)
\(=1+1+1+1+...+1\) (có \(51\) số \(1\))
\(=51\)
Tính \(f\left(-1\right)\)
\(f\left(x\right)=1+x^3+x^5+x^7+...+x^{101}\)
\(\Rightarrow f\left(-1\right)=1+\left(-1\right)^3+\left(-1\right)^5+...+\left(-1\right)^{101}\)
\(=1+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\) (có \(50\) số \(-1\))
\(=1+\left(-50\right)\)
\(=-49\)
Vậy: \(\left\{{}\begin{matrix}f\left(1\right)=51\\f\left(-1\right)=-49\end{matrix}\right.\)
Ta có:
a) \(f\left(1\right)=1+1^3+1^5+1^7+...+1^{101}\)
\(f\left(1\right)=1+50=51\)
b) \(f\left(-1\right)=1+\left(-1\right)^3+\left(-1\right)^5+\left(-1\right)^7+...+\left(-1\right)^{101}\)
\(f\left(-1\right)=1-50=-49\)