数列{an}满足 a1=1 a(n+1)=2an-n^2+3n 求an通项公式!_数学解答

数列{an}满足 a1=1 a(n+1)=2an-n^2+3n 求an通项公式!

人气：222 ℃　时间：2020-04-06 04:54:13

解答

用递推法a2-a1=a1-1*1+3*1a3-a2=a2-2*2+3*2a4-a3=a3-3*3+3*4.an-a(n-1)=2a(n-1)-(n-1)*(n-)+3*(n-1)等号左右分别相加得an+an-a1=an+S(n-1)-[1*1+2*2+3*3..(n-1.*(n-1)]+3[1+2+3..(n-1)]=Sn-[1*1+2*2+3*3..(n-1.*(n-1...