Question

\(\left\{{}\begin{matrix}U_n=1\\U_{n+1}=U_n+\left(n-1\right)2^n\end{matrix}\right.\)

a) tìm công thức tổng quát

b) cm dãy sỗ tăng

Answer 1

\(u_{n+1}=u_n+\left(n-1\right)2^n\)

\(\Leftrightarrow u_{n+1}-\left(n+1\right)2^{n+1}+3.2^{n+1}=u_n-n.2^n+3.2^n\)

Đặt \(v_n=u_n-n.2^n+3.2^n\Rightarrow\left\{{}\begin{matrix}v_1=5\\v_{n+1}=v_n=...=v_1=5\end{matrix}\right.\)

\(\Rightarrow u_n-n.2^n+3.2^n=5\)

\(\Rightarrow u_n=\left(n-3\right)2^n+5\)

b. ta có:

\(u_{n+1}-u_n=\left(n-1\right)2^n\ge0\) ; \(\forall n\ge1\)

\(\Rightarrow u_{n+1}\ge u_n\Rightarrow\) dãy tăng