已知数列的通项公式Bn=1/n(n+2),求数列的前n项和.？
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Bn=1/n(n+2)=(1/2)[1/n-1/(n+2)]
Sn=(1/2)[1/1-1/3+1/2-1/4+1/3-1/5+...+1/(n-1)-1/(n+1)+1/n-1(n+2)]
=(1/2)[1/1+1/3-1/(n+1)-1/(n+2)]
裂项求和.,9,s1=1/1*3 =1/2(1-1/3) s2=1/2*4 =1/2(1/2-1/4) s3=1/2(1/3-1/5)
sn=1/2{1-1/3+1/2-1/4+1/3-1/5+---------+1/n-1/(n+2)}=3/4+1/2(n+1)(n+2),0,
Sn=(1/2)[1/1-1/3+1/2-1/4+1/3-1/5+...+1/(n-1)-1/(n+1)+1/n-1(n+2)]
=(1/2)[1/1+1/3-1/(n+1)-1/(n+2)]
裂项求和.,9,s1=1/1*3 =1/2(1-1/3) s2=1/2*4 =1/2(1/2-1/4) s3=1/2(1/3-1/5)
sn=1/2{1-1/3+1/2-1/4+1/3-1/5+---------+1/n-1/(n+2)}=3/4+1/2(n+1)(n+2),0,
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