\( a_{n+1}^2+a_{n}^2+4=2a_{n+1}a_{n}+4a_{n+1}+4a_{n} \)......(1)式
\( a_{n}^2+a_{n-1}^2+4=2a_{n}a_{n-1}+4a_{n}+4a_{n-1} \)......(2)式
(1)式-(2)式得
\( a_{n+1}^2-a_{n-1}^2=2a_{n}(a_{n+1}-a_{n-1})+4(a_{n+1}-a_{n-1}) \)
\( (a_{n+1}+a_{n-1})(a_{n+1}-a_{n-1})-2a_{n}(a_{n+1}-a_{n-1})-4(a_{n+1}-a_{n-1})=0 \)
\( (a_{n+1}-a_{n-1})(a_{n+1}+a_{n-1}-2a_{n}-4)=0 \)
得到\( a_{n+1}+a_{n-1}-2a_{n}-4=0 \),\( (a_{n+1}-a_{n})-(a_{n}-a_{n-1})=4 \)
令\( b_n=a_{n}-a_{n-1} \),\( n>1 \),\( b_2=a_{2}-a_{1}=8-2=6 \)
列出遞迴式
\( b_{n}-b_{n-1}=4 \)
\( b_{n-1}-b_{n-2}=4 \)
...
\( b_{3}-b_{2}=4 \)
以上的式子相加\( b_{n}-b_{2}=4(n-2) \),\( b_{n}=4n-2 \)
列出遞迴式
\( a_{n}-a_{n-1}=4n-2 \)
\( a_{n-1}-a_{n-2}=4(n-1)-2 \)
...
\( a_{2}-a_{1}=4*2-2 \)
以上的式子相加\( a_{n}-a_{1}=\frac{n-1}{2}(4n-2+6) \),\( a_{n}=2n^2 \)
111.4.19補充
設一數列\(\langle a_n \rangle\)滿足\(a_1=1\),\(a_{n+1}>a_n(n\in N)\)且\((a_{n+1})^2+(a_n)^2+1=2(a_{n+1}\cdot a_n+a_{n+1}+a_n)\)。令\(\displaystyle S_n=\sum_{k=1}^n a_k\),試求\(\displaystyle \lim_{n\to \infty}\frac{S_n}{na_n}=\)
。
(111台中一中,
https://math.pro/db/viewthread.php?tid=3621&page=1#pid23757)