Question

Cho dãy số \(\left(u_n\right)\) xác định bởi \(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=u_n+n^3,\forall n\in N^{\cdot}\end{matrix}\right.\) . Tìm số nguyên dương n nhỏ nhất sao cho \(\sqrt{u_n-1}\ge2039190\)

Answer 1

Ta co :

\(U_2=U_1+1^3\)

\(U_3=U_2+2^3\)

...

\(U_n=U_{n-1}+\left(n-1\right)^3\)

cong ve :

\(\Rightarrow U_n=U_1+1^3+2^2+...+\left(n-1\right)^3\)

\(=1+\left(\frac{n.\left(n-1\right)}{2}\right)^2\)

\(\Rightarrow\sqrt{U_n-1}=\sqrt{1+\left(\frac{n.\left(n-1\right)}{2}\right)^2-1}=\frac{n.\left(n-1\right)}{2}\ge2039190\)

\(\Rightarrow n^2-n\ge4078380\)

\(\Rightarrow\left[{}\begin{matrix}n\le-2019\\n\ge2020\end{matrix}\right.\)(n ϵ N*)\(\Rightarrow n\ge2020\)