已知数列{an}满足a1=1,an+1=an+1/n(n+1),写出前五项,并归纳出数列的一个通项公式
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![已知数列{an}满足a1=1,an+1=an+1/n(n+1),写出前五项,并归纳出数列的一个通项公式](/uploads/image/z/164320-16-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%2B1%3Dan%2B1%2Fn%28n%2B1%29%2C%E5%86%99%E5%87%BA%E5%89%8D%E4%BA%94%E9%A1%B9%2C%E5%B9%B6%E5%BD%92%E7%BA%B3%E5%87%BA%E6%95%B0%E5%88%97%E7%9A%84%E4%B8%80%E4%B8%AA%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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已知数列{an}满足a1=1,an+1=an+1/n(n+1),写出前五项,并归纳出数列的一个通项公式
已知数列{an}满足a1=1,an+1=an+1/n(n+1),写出前五项,并归纳出数列的一个通项公式
已知数列{an}满足a1=1,an+1=an+1/n(n+1),写出前五项,并归纳出数列的一个通项公式
a(n+1)=an+1/n(n+1)
a1=1
a2=a1+1/1*2
=1+1/2
=3/2
a3=a2+1/2*3
=3/2+1/6
=5/3
a4=a3+1/3*4
=5/3+1/12
=7/4
a5=a4+1/4*5
=7/4+1/20
=36/20
=9/5
a(n+1)=an+1/n(n+1)
a(n+1)-an=1/n(n+1)
an-a(n-1)=1/(n-1)n
.
a3-a2=1/2*3
a2-a1=1/1*2
以上等式相加得
an-a1=1/1*2+1/2*3+1/3*4+.+1/n*(n+1)
an-a1=1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
an-a1=1-1/(n+1)
an-1=n/(n+1)
an=(2n+1)/(n+1)
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