Đáp án:

\[\lim {u_n} = \frac{5}{3}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
{u_{n + 2}} = \frac{{{u_{n + 1}} + {u_n}}}{2}\\
 \Leftrightarrow 2{u_{n + 2}} = {u_{n + 1}} + {u_n}\\
 \Leftrightarrow 2{u_{n + 2}} - 2{u_{n + 1}} =  - {u_{n + 1}} + {u_n}\\
 \Leftrightarrow {u_{n + 2}} - {u_{n + 1}} = \frac{{ - 1}}{2}\left( {{u_{n + 1}} - {u_n}} \right)\\
{x_n} = {u_{n + 1}} - {u_n} \Rightarrow \left\{ \begin{array}{l}
{x_{n + 1}} =  - \frac{1}{2}{x_n}\\
{x_1} = {u_2} - {u_1} = 1
\end{array} \right.\\
 \Rightarrow {x_n} = {x_1}.{\left( { - \frac{1}{2}} \right)^{n - 1}} = {\left( { - \frac{1}{2}} \right)^{n - 1}}\\
 \Rightarrow \left\{ \begin{array}{l}
{u_2} - {u_1} = 1\\
{u_3} - {u_2} = \left( { - \frac{1}{2}} \right)\\
{u_4} - {u_3} = {\left( { - \frac{1}{2}} \right)^2}\\
....\\
{u_n} - {u_{n - 1}} = {\left( { - \frac{1}{2}} \right)^{n - 2}}
\end{array} \right.\\
 \Rightarrow \left( {{u_2} - {u_1}} \right) + \left( {{u_3} - {u_2}} \right) + \left( {{u_4} - {u_3}} \right) + .... + \left( {{u_n} - {u_{n - 1}}} \right) = 1 + \left( { - \frac{1}{2}} \right) + {\left( { - \frac{1}{2}} \right)^2} + .... + {\left( { - \frac{1}{2}} \right)^{n - 2}}\\
 \Leftrightarrow {u_n} - {u_1} = \frac{{1 - {{\left( {\frac{{ - 1}}{2}} \right)}^{n - 1}}}}{{1 - \left( { - \frac{1}{2}} \right)}}\\
 \Rightarrow {u_n} = 1 + \frac{{1 - {{\left( {\frac{{ - 1}}{2}} \right)}^{n - 1}}}}{{\frac{3}{2}}}\\
 \Rightarrow \lim {u_n} = 1 + \frac{1}{{\frac{3}{2}}} = \frac{5}{3}
\end{array}\)

Ta có:

un+2=un+1+un2⇔2un+2=un+1+un⇔2un+2−2un+1=−un+1+un⇔un+2−un+1=−12(un+1−un)xn=un+1−un⇒{xn+1=−12xnx1=u2−u1=1⇒xn=x1.(−12)n−1=(−12)n−1⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩u2−u1=1u3−u2=(−12)u4−u3=(−12)2....un−un−1=(−12)n−2⇒(u2−u1)+(u3−u2)+(u4−u3)+....+(un−un−1)=1+(−12)+(−12)2+....+(−12)n−2⇔un−u1=1−(−12)n−11−(−12)⇒un=1+1−(−12)n−132⇒limun=1+132=53