级数2/(n+2)(n+1)n 怎么求出答案
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级数2/(n+2)(n+1)n 怎么求出答案
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2/[n(n+1)(n+2)] = A/n+B/(n+1)+C/(n+2)
= [A(n+1)(n+2)+Bn(n+2)+Cn(n+1)]/[n(n+1)(n+2)]
= [(A+B+C)n^2+(3A+2B+C)n+2A]/[n(n+1)(n+2)] ,
比较n的同次幂的系数,得
A+B+C=0
3A+2B+C=0
2A=2
联立解得 A=1,B=-2.C=1
则 2/[n(n+1)(n+2)] = 1/n-2/(n+1)+1/(n+2),
∑ 2/[n(n+1)(n+2)]
= ∑[1/n-2/(n+1)+1/(n+2)]
=1-2/2+1/3 +1/2-2/3+1/4 +1/3-2/4+1/5 +.= 1/2.
= [A(n+1)(n+2)+Bn(n+2)+Cn(n+1)]/[n(n+1)(n+2)]
= [(A+B+C)n^2+(3A+2B+C)n+2A]/[n(n+1)(n+2)] ,
比较n的同次幂的系数,得
A+B+C=0
3A+2B+C=0
2A=2
联立解得 A=1,B=-2.C=1
则 2/[n(n+1)(n+2)] = 1/n-2/(n+1)+1/(n+2),
∑ 2/[n(n+1)(n+2)]
= ∑[1/n-2/(n+1)+1/(n+2)]
=1-2/2+1/3 +1/2-2/3+1/4 +1/3-2/4+1/5 +.= 1/2.