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一、利用公式C0n+C1n+C2n+C3n+…+Cn n=2n求和1.直接利用公式例1求和C1n+C3n+C5n+…解由于奇数项之和与偶数项之和相等,因此奇数项之和等于所有项之和的一半.所以C1n+C3n+C5n+…=1/2×2n=2n-1.2.由公式Cr n=Cn-r n进行转化例2求和1+2C1n+3C2n+…+(n+1)Cn n.解设S=1+2C1n+3C2n+…+(n+1)Cn n,其倒序和为S=(n+1)Cn n+nCn-1n+…+2C1n+1.考虑到Cr n=Cn-r n(0≤r≤n),将以上两式相加得2S=(n+2)C0n+(n+2)C1n+…+(n+2)Cn n=(n+2)·2n,所以S=(n+2)·2n-1
First, the use of the formula C0n + C1n + C2n + C3n + ... + Cn n = 2n summation 1. Direct use of the formula Example 1 summing C1n + C3n + C5n + ... Solution Since the sum of odd and even terms and the sum of the odd terms So that C1n + C3n + C5n + ... = 1/2 × 2n = 2n-1.2 Conversion by the formula Cr n = Cn-r n Example 2 Sum 1 + 2C1n + 3C2n + ... + (n + 1) Cn n. Solving S = 1 + 2C1n + 3C2n + ... + (n + 1) Cn n, the reverse order is S = (n + 1) Cn n + nCn- 1n + ... + 2C1n + 1. Considering that Cr n = Cn-r n (0 ≦ r ≦ n), the above two equations are summed so that 2S = n + 2 C 0n + n + 2 C n + +2) · 2n, S = (n + 2) · 2n-1