数学基础之数列求和

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错位相减

证明等比数列求和 : ∑k=0napk=a1−pn+11−p\sum_{k=0}^{n}ap^{k} = a\frac{1-p^{n+1}}{1-p}

证明

Sn=∑k=0napk=a∑k=0npkpSn=a∑k=0npk+1=a∑k=1n+1pk(1−p)Sn=a(1−pn+1)Sn=a1−pn+11−pS_n =\sum_{k=0}^{n}ap^{k} = a\sum_{k=0}^{n}p^k\\ pS_n = a\sum_{k=0}^{n}p^{k+1} = a\sum_{k=1}^{n+1}p^{k}\\ (1-p)S_n = a(1 - p^{n+1})\\ S_n = a\frac{1 - p^{n+1}}{1-p}

更一般的, 化简 : ∑k=0nkpk\sum_{k=0}^{n}kp^k

Sn=∑k=0nkpkpSn=∑k=0nkpk+1=∑k=1n+1(k−1)pk(1−p)Sn=∑k=0nkpk−∑k=1n+1(k−1)pk(1−p)Sn=∑k=1nkpk−npn+1−∑k=1n(k−1)pk(1−p)Sn=∑k=1npk−npn+1(1−p)Sn=p1−pn1−p−npn+1Sn=11−p(p1−pn1−p−npn+1)\begin{aligned} S_n = \sum_{k=0}^{n}kp^k\\ pS_n = \sum_{k=0}^{n}kp^{k+1} = \sum_{k=1}^{n + 1}(k-1)p^{k}\\ (1-p)S_n = \sum_{k=0}^{n}kp^k - \sum_{k=1}^{n + 1}(k-1)p^{k}\\ (1-p)S_n = \sum_{k=1}^{n}kp^k - np^{n + 1} - \sum_{k=1}^{n}(k-1)p^{k}\\ (1-p)S_n = \sum_{k=1}^{n}p^k - np^{n + 1}\\ (1-p)S_n = p\frac{1-p^n}{1-p} - np^{n + 1}\\ S_n = \frac{1}{1-p}(p\frac{1-p^n}{1-p} - np^{n + 1})\\ \end{aligned}

裂项相消

化简 Sn=∑k=1n(ak+b)qkS_n = \sum_{k=1}^{n}(ak + b)q^k

设(ak+b)qk=f(k+1)qk+1−f(k)qk,f(k)=Ak+B(ak+b)qk=(qf(k+1)−f(k))qkak+b=qf(k+1)−f(k)=qA(k+1)+qB−Ak−Bak+b=qf(k+1)−f(k)=(q−1)Ak+qA+(q−1)BA=aq−1,B=b−qAq+1=b−qaq−1q−1f(x)已知∑k=1nak=f(n+1)qn+1−f(1)q设\quad (ak+b)q^k = f(k + 1)q^{k+1} - f(k)q^{k} , f(k) = Ak + B\\ (ak+b)q^k = (qf(k + 1) - f(k))q^{k}\\ ak+b = qf(k+1) - f(k) = qA(k+1) + qB - Ak - B\\ ak+b = qf(k+1) - f(k) = (q-1)Ak + qA + (q - 1)B\\ A = \frac{a}{q-1}, B = \frac{b - qA}{q+1} = \frac{b - q\frac{a}{q-1}}{q-1}\\ f(x) 已知\\ \sum_{k=1}^{n}a_k = f(n+1)q^{n + 1} - f(1)q\\