设数列{an}的通项公式an=n·2^(n-1)求数列前n项和Sn(用错位相减法最好)
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![设数列{an}的通项公式an=n·2^(n-1)求数列前n项和Sn(用错位相减法最好)](/uploads/image/z/2708638-70-8.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3Dn%C2%B72%5E%EF%BC%88n-1%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%EF%BC%88%E7%94%A8%E9%94%99%E4%BD%8D%E7%9B%B8%E5%87%8F%E6%B3%95%E6%9C%80%E5%A5%BD%EF%BC%89)
设数列{an}的通项公式an=n·2^(n-1)求数列前n项和Sn(用错位相减法最好)
设数列{an}的通项公式an=n·2^(n-1)求数列前n项和Sn(用错位相减法最好)
设数列{an}的通项公式an=n·2^(n-1)求数列前n项和Sn(用错位相减法最好)
Sn= 1*2^0 +2*2^1 +3*2^2 +4*2^3 +...+n*2^(n-1)
2Sn= 1*2^1 +2*2^2 +3*2^3 +4*2^4 +...+n*2^n
所以 2Sn-Sn= -1*2^0 +(1-2)*2^1 +(2-3)*2^2 +(3-4)*2^3 +...+(n-1-n)*2^(n-1) +n*2^n
所以 Sn= -1 -[2^1 +2^2 +2^3 +...+2^(n-1)] +n*2^n
=n*2^n -1 +2 -2^n
=(n-1)*2^n +1