留个记号,我也想知道着么写无限。
PI = 4*(1 – 1/3 +1/5 – 1/7 +1/9 – 1/11 + 1/13 + … )这个数学式如何用j
3个回答
相关问题
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1/1*3*5+1/3*5*7+1/5*7*9+1/7*9*11+1/9*11*13+1/11*13*15怎么计算
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1+3 1+3+5 1+3+5+7 1+3+5+7+9 1+3+5+7+9+11 1+3+5+7+9+11+13 1+3
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(1\5+1\7+9\1+1\11)x(1\7+1\9+1\11+1\13)-(1\5+1\7+1\9+1\11+1\1
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int j,a[]={1,3,5,7,9,11,13,15,
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1+3+5+7+9+11+13+15+13+11+9+7+5+3+1=113=
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1.1+3.3+5.5+7.7+9.9+11.11+13.13+15.15
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进来帮我看看计算1/(1*3)+1/(3*5)+1/(5*7)+1/(7*9)+1/(9*11)+1/(11*13)
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1/5*7+1/7*9+1/9*11+1/11*13+.+1/23*25
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1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13=
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(1-1/3+1/5-1/7+1/9-1/11+1/13-1/15+1/17)×4