1+1/2+1/4+1/8+...1/64=?
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1+1/2+1/4+1/8+...1/64=?1+1/2+1/4+1/8+...1/64=?1+1/2+1/4+1/8+...1/64=?就是一个等比数列的求和问题.Sn=a1(1-q^n)/(1-q
1+1/2+1/4+1/8+...1/64=?
1+1/2+1/4+1/8+...1/64=?
1+1/2+1/4+1/8+...1/64=?
就是一个等比数列的求和问题.
Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1)
=(1 - 1/64 · 1/2)/(1-1/2)
=127/64.
可以这样计算
1+1/2+1/4+1/8+1/16+1/32+1/64
=1+1/2+1/4+1/8+1/16+1/32+(1/64+1/64)-1/64
=1+1/2+1/4+1/8+1/16+(1/32+1/32)-1/64
=1+1/2+1/4+1/8+(1/16+1/16)-1/64
=1+1-1/64
=1+(63/64)
所以选择C
1/2+1/4+1/6+1/8+.1/n
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