问题 填空题
定义运算:∏limit
sni=1
ai=a1a2a3•…•an
,已知ai=logi+1(i+2),计算∏limit
s62i=1
ai
=______.
答案

∵ai=logi+1(i+2)=

lg(i+2)
lg(i+1)
πlimit
sni=1
ai=a1a2a3•…•an
=
lg3
lg2
lg4
lg3
lg(n+2)
lg(n+1)

πlimit

s62i=1
ai=
lg3
lg2
lg4
lg3
lg64
lg63
=
lg64
lg2
=
6lg2
lg2
=6.

故答案为:6.

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