Bài 1:
\(x^2+x^4+x^6+x^8+...+x^{100}\)
Thay \(x=1\) ta có:
\(1^2+1^4+1^6+1^8+...+1^{100}\)
\(=1+1+1+1+...+1\) (có \(50\) số \(1\))
\(=1.50\)
\(=50\)
Thay \(x=-1\) ta có:
\(\left(-1\right)^2+\left(-1\right)^4+\left(-1\right)^6+...+\left(-1\right)^{100}\)
\(=1+1+1+...+1\) (có \(50\) số \(1\))
\(=1.50\)
Bài 2:
\(f\left(x\right)=1+x^3+x^5+x^7+...+x^{101}\)
Tính \(f\left(1\right):\)
\(f\left(1\right)=1+1^3+1^5+1^7+...+1^{101}\)
\(\Rightarrow f\left(1\right)=1+1+1+1+...+1\) (có \(51\) số \(1\))
\(\Rightarrow f\left(1\right)=1.51\)
\(\Rightarrow f\left(1\right)=51\)
Tính \(f\left(-1\right):\)
\(f\left(-1\right)=1+\left(-1\right)^3+...+\left(-1\right)^{101}\)
\(\Rightarrow f\left(-1\right)=1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(có \(50\) số \(-1\))
\(\Rightarrow f\left(-1\right)=1+1.\left(-50\right)\)
\(\Rightarrow f\left(-1\right)=1+\left(-50\right)\)
\(\Rightarrow f\left(-1\right)=-49\)
Vậy: \(\left\{{}\begin{matrix}f\left(1\right)=51\\f\left(-1\right)=-49\end{matrix}\right.\)
Câu 1:
Tại x=1 thì x2 + x4 + x6 + x8 + ... + x100 =50
Tại x=-1 thì x2 + x4 + x6 + x8 + ... + x100 =50
Câu 2:
\(f\left(1\right)=51\)
\(f\left(-1\right)=-49\)
