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) =(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
|1/4-1/2|+|1/6-1/4|+|1/8-1/6|+```+|1/2008-1/2006|
((1/2)-1)*((1/3)-1)*((1/4)-1)*((1/5)-1)*((1/6)-1)*((1/7)-1)*((1/8)-1)*((1/9)-1)*((1/10)-1)
1+1+1+1+1+1+1+1+1+2+5+4+8+3+6+2+1+4等于多少?
2/1+4/1+4/1+6/1+8/1+16/1+32/1=?
9*(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*(1-1/6)*(1-1/7)*(1-1/8)*(1-1/9)怎样简便计算
1/2*4+1/4*6+1/6*8+1/8*10+---+1/198*200
4,2,1,1/2,1/4,( ) ( ) ( )1/8,1/32,1/4 ,1/16( )( ) ( )1/2,1/6,1/12,1/20,1/30,1/42 ( ) ( ) ( )
2/1 3/1 4/1 5/1 6/1 8/1 10/1 12/1 15/1 20/1
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1+1+2+1+2+6+4+5+8等于
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计算:|1/4-1/2|+|1/6-1/4|+|1/8-1/6|+…+|1/2008-1/2006|1003/2008
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