等差数列求和,每三项合并,有668项,就可以用一加二一直加到一百的方法算了
m-2m-3m+4m-5m-6m+7m-8m-9m+.+2002m-2003m-2004m
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相关问题
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m-2m-3m+4m-5m-6m+7m.2004m=?
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2m+3m+5m+7m+9m+…+2003m+2005m+2007m
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(m+3m5m+7m+.2009m)-(2m+4m+6m.2008m)
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(m+3m+5m+7m+.+2009m)-(2m-4m-6m-.-2008m)=?
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A=M0∧M3∧M5. ∴┑A=M1∧M2∧M4∧M6∧M7.
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已知m=2005,求m/(1x2)+m/(2x3)+m/(3x4)...+m/(2003x2004)+m/(2004x2
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计算(3m^2)^3-7m^3【m^3-m(4m^2+1)】等于 A.7m^3+9m^4 B.7m^4+48m^6 C.
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(m+3m+5m+...+2009m)-(2m+4m+6m+...2008m)
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(m+3m+5m+...+2015m)-(2m+4m+6m+...+2014m)=
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计算:(m+3m+5m+……2013m)-(2m+4m+6m+…2012m)=_