数列(an)满足啊a1+2a2+3a3+4a4+……+nan=n(n+1)(n+2)(n≥2且n为自然数)求an的通项公式.
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数列(an)满足啊a1+2a2+3a3+4a4+……+nan=n(n+1)(n+2)(n≥2且n为自然数)求an的通项公式.
数列(an)满足啊a1+2a2+3a3+4a4+……+nan=n(n+1)(n+2)(n≥2且n为自然数)求an的通项公式.
数列(an)满足啊a1+2a2+3a3+4a4+……+nan=n(n+1)(n+2)(n≥2且n为自然数)求an的通项公式.
an=3n+3,你把a1 ,a2 ,a3,a4 算出来,会观察到他们成等差数列,于是猜想an是等差数列,再用数学归纳法证明,不知你是否学过数学归纳法.可以看看.还有个很简单的方法就是a1+2a2+3a3+……+nan =n(n+1)(n+2),a1+2a2+3a3+……(n-1)an-1 =(n-1)n(n+1),这个式子应该知道怎么来的吧.然后把前一个式子减后一个式子就可以得出an=3n+3,这种两式相减得通项在数列里是很常见的哦.
a1+2a2+3a3+4a4+……+(n-1)a(n-1)=(n-1)n(n+1)
a1+2a2+3a3+4a4+……+ nan = n(n+1)(n+2)
两式相减,得an=3n+3
a1+2a2+.....+nan=n(n+1)(n+2)
a1+2a2+.....+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)
两个式子想减,可得答案
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