本站别名http://im.cx/n
a1=2,a2=1/2,且[3+(-1)^n]A(n+2)-2An+2[(-1)^n-1]=0,n属于自然数，求An通项公式 a1=2,a2=1/2,a3=4,a4=1/4,a5=6,a6=1/8 奇数项部分是an=n+1, 偶数项部分是an=2^(-n/2);
则通项公式是：
![\[ \begin{array}{l} A_n = \frac{{\left( { - 1} \right)^{n - 1} + 1}}{2}\left( {n + 1} \right) + \frac{{\left( { - 1} \right)^n + 1}}{2}2^{ - \frac{n}{2}} \begin{array}{*{20}c} {} & {} \\ \end{array}or \\ A_n = \left( {n - 2\left\lfloor {\frac{n}{2}} \right\rfloor } \right)\left( {n + 1} \right) + \left( {n + 1 - 2\left\lfloor {\frac{{n + 1}}{2}} \right\rfloor } \right)2^{ - \frac{n}{2}} \\ \end{array} \]](http://lucideye.is-programmer.com/user_files/lucideye/epics/1c48e5b1f968b3fdc7a7316eeefeb1c0b04ba5ea.png "\[ \begin{array}{l} A_n = \frac{{\left( { - 1} \right)^{n - 1} + 1}}{2}\left( {n + 1} \right) + \frac{{\left( { - 1} \right)^n + 1}}{2}2^{ - \frac{n}{2}} \begin{array}{*{20}c} {} & {} \\ \end{array}or \\ A_n = \left( {n - 2\left\lfloor {\frac{n}{2}} \right\rfloor } \right)\left( {n + 1} \right) + \left( {n + 1 - 2\left\lfloor {\frac{{n + 1}}{2}} \right\rfloor } \right)2^{ - \frac{n}{2}} \\ \end{array} \]")
评论 (0)