C4 递推关系与生成函数
S0 斐波那契数列
1)递推公式:fn+2=fn+1+fn,f0=0,f1=1f_{n+2} = f_{n+1}+f_n,f_0 = 0,f_1 = 1fn+2=fn+1+fn,f0=0,f1=1
2)通项公式:fn=15[(1+52)n−(1−52)n]f_n = \frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n]fn=51[(21+5)n−(21−5)n]
3)与二项式系数关系:fn=∑k=0n−1Cn−1−kkf_n = \sum\limits_{k=0}^{n-1}C_{n-1-k}^kfn=k=0∑n−1Cn−1−kk,即 Pascal 三角形从左下到右上每条线的和
4)部分和:Sn=fn+2−1S_n = f_{n+2}-1Sn=fn+2−1
S1 生成函数
1)生成函数:f(x)=∑k=0∞akxkf(x) = \sum\limits_{k=0}^\infin a_k x^kf(x)=k=0∑∞akxk
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Sn=∑k=0nakS_n=\sum\limits_{k=0}^na_kSn=k=0∑nak 的生成函数:f(x)1−x\frac{f(x)}{1-x}1−xf(x)
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kkk 元无限多重集 nnn 组合(∑i=1kcixi=n\sum\limits_{i=1}^k c_ix_i = ni=1∑kci