原式=(1+1)^n =2^n
原式=(1+1)^n =2^n
组合公式Cmn中m取n所有值之和Cnn+Cn-1n+...+C1n+C0n=? 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 原式=(1+1)^n =2^n 解析看不懂?免费查看同类题视频解析查看解答 相似问题 VB.NET2010 给定求组合数公式为Cmn=m!/n!(m-n)!,编一程序,输入m和n的值,求Cmn的值.要...
则 C 2n+3 = C 2n+1 + C 1n+1 + C 2n 即 (n+3)(n+2) 2 = (n+1)n 2 +(n+1)+ n(n-1) 2 , 整理得n 2 -3n-4=0 又由n∈N * ,则
In Proceedings of the 2018 International Conference on Advances in Computing, Communication Control and Networking (ICACCCN), Greater Noida (UP), India, 12–13 October 2018; pp. 142–145. 18. Malik, H.; Roy, N. Extreme Learning Machine-Based Image Classification Model Using Handwritten Digit...
=Cn+1n+m+Cnn+m =Cn+1n+m+1. 本题考查了组合数的计算,需要学生熟练掌握组合数的性质; 首先由Cnn=Cn+1n+1,Cn+1n+1+Cnn+1=Cn+1n+2,逐步对其进行化简; 由上述提示即得Cnn+Cnn+1+Cnn+2+…+Cnn+m=Cn+1n+m+1=Cn+1n+m+Cnn+m,进一步利用组合数的性质化简,即可证明结论.结果...
解析 以下等式成立: Cnn=Cn+1n+1, Cn+1n+1+Cnn+1=Cn+1n+2, Cn+1n+2+Cnn+2=Cn+1n+3, ... Cn+1n+m+Cnn+m=Cn+1n+m+1 把以上式子相加即可得出 Cnn+Cnn+1+Cnn+2+...+Cnn+m=Cn+1n+m+1 可以根据组合数的性质得出答案。
∵(1+x)n×(x+1)n=(1+x)n×(1+x)n=(1+x)2n,取左右两边展开式中含xn-1项的系数,则(C0n•C1n+C1n•C2n+…+Cn-1n•Cnn)•xn-1=Cn-12n•xn-1,∴C0n•C1n+C1n•C2n+…+Cn-1n•Cnn=Cn-12n=(2n)!(n-1)!•(n+... 相关...
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Cn2n=Cn+12n+1. 相关知识点: 试题来源: 解析 证明:∵Cm+1n+Cmn=Cm+1n+1,∴Cnn+Cnn+1+…+Cn2n−1+Cn2n=(Cn+1n+1+Cnn+1)+Cnn+2+…+Cn2n−1+Cn2n=Cn+1n+2+Cnn+2+…+Cn2n−1+Cn2n=Cn+1n+3+…+Cn2n−1+Cn2n=…=Cn+12n−1+Cn2n−1+Cn2n=Cn+12n+Cn2n=Cn+12n+1....