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已知数列{an}:an=8/((n+1)(n+3)),求(n+1)(an-an+1)和的值.
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已知数列{an}:an=8/((n+1)(n+3)),求(n+1)(an-an+1)和的值.
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答案和解析
an=8/((n+1)(n+3))=4（1/(n+1)-1/(n+3))
令bn=(n+1)(an-an+1)
bn=(n+1)(an-an+1)
=4(n+1)[(1/(n+1)-1/(n+3))-(1/(n+2)-1/(n+4))]
=4(n+1)[1/(n+1)-1/(n+2)-1/(n+3)+1/(n+4)]
=4[1-(1-1/(n+2))-(1-2/(n+3))+(1-3/(n+4))]
=4[1/(n+2)+2/(n+3)-3/(n+4)]
那么bn前n项和：
Tn=4[(1/3+2/4-3/5)+(1/4+2/5-3/6)+(1/5+2/6-3/7)+...+1/(n+2)+2/(n+3)-3/(n+4)]
=4[1/3+2/4+1/4-1/(n+3)-3/(n+4)]
=13/3-4/(n+3)-12/(n+4)
令bn=(n+1)(an-an+1)
bn=(n+1)(an-an+1)
=4(n+1)[(1/(n+1)-1/(n+3))-(1/(n+2)-1/(n+4))]
=4(n+1)[1/(n+1)-1/(n+2)-1/(n+3)+1/(n+4)]
=4[1-(1-1/(n+2))-(1-2/(n+3))+(1-3/(n+4))]
=4[1/(n+2)+2/(n+3)-3/(n+4)]
那么bn前n项和：
Tn=4[(1/3+2/4-3/5)+(1/4+2/5-3/6)+(1/5+2/6-3/7)+...+1/(n+2)+2/(n+3)-3/(n+4)]
=4[1/3+2/4+1/4-1/(n+3)-3/(n+4)]
=13/3-4/(n+3)-12/(n+4)
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