求:Cn0+3Cn1+5Cn2+…+(2n+1)Cnn=s 求s(注:各项均为二项式的项:n在下;0、1、2、…n在上;3、5、…(2n+1)为系数、求详解、谢谢!)答案再此设:S=Cn^0+3Cn^1+5Cn^2+…+(2n+1)Cn^n S=(2n+1)Cn^n+(2
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![求:Cn0+3Cn1+5Cn2+…+(2n+1)Cnn=s 求s(注:各项均为二项式的项:n在下;0、1、2、…n在上;3、5、…(2n+1)为系数、求详解、谢谢!)答案再此设:S=Cn^0+3Cn^1+5Cn^2+…+(2n+1)Cn^n S=(2n+1)Cn^n+(2](/uploads/image/z/11489074-34-4.jpg?t=%E6%B1%82%EF%BC%9ACn0%2B3Cn1%2B5Cn2%2B%E2%80%A6%2B%EF%BC%882n%2B1%EF%BC%89Cnn%3Ds+%E6%B1%82s%EF%BC%88%E6%B3%A8%EF%BC%9A%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E4%BA%8C%E9%A1%B9%E5%BC%8F%E7%9A%84%E9%A1%B9%EF%BC%9An%E5%9C%A8%E4%B8%8B%EF%BC%9B0%E3%80%811%E3%80%812%E3%80%81%E2%80%A6n%E5%9C%A8%E4%B8%8A%EF%BC%9B3%E3%80%815%E3%80%81%E2%80%A6%282n%2B1%29%E4%B8%BA%E7%B3%BB%E6%95%B0%E3%80%81%E6%B1%82%E8%AF%A6%E8%A7%A3%E3%80%81%E8%B0%A2%E8%B0%A2%21%EF%BC%89%E7%AD%94%E6%A1%88%E5%86%8D%E6%AD%A4%E8%AE%BE%EF%BC%9AS%3DCn%5E0%EF%BC%8B3Cn%5E1%EF%BC%8B5Cn%5E2%EF%BC%8B%E2%80%A6%EF%BC%8B%282n%EF%BC%8B1%29Cn%5En+S%3D%282n%EF%BC%8B1%29Cn%5En%EF%BC%8B%282)
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