急问高一函数
若 a1=1 a2=1/2 a3=3/4 a4 =5/8 a5 =11/6 ····· 求an的通项公式 ~~ 感谢~~
A2-A1 = -1/2;A3-A2 = 1/4;A4-A3 = -1/8;A5-A4=1/16 ..... An -A(n-1) = (-1)^n/2^(n-1) 上面式子,二边加和,得: An -A1 = (-1/2)+(1/4)+(-1/8)+...+(-1)^n/2^(n-1) 右边为首项为-1/2、公比为-1/2的等比数列。 因此可得,An = A1 + (-1/2)[(-1/2)^(n-1)-1]/[(-1/2)-1] An = [2 +(-2)^(1-n)]/3
A2-A1 = -1/2 A3-A2 = 1/4 A4-A3 = -1/8 A5-A4=1/16 ..... An -A(n-1) = (-1)^n/2^(n-1) (+) -------------------------------------------------- An -A1 = (-1/2)+(1/4)+(-1/8)+...+(-1)^n/2^(n-1) An = A1 + (-1/2)[(-1/2)^(n-1)-1]/[(-1/2)-1] An =1+[(-1/2)^(n-1)-1]/3
答:数列3,5,7,15,... 的一个通项公式为: An=(2n+1)+(n-1)(n-2)(n-3),(n∈N*)详情>>
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