阿贝尔变换

阿贝尔(Abel)变换

什么是 Abel 变换

设 \(B_{n} = \sum\limits_{i = 1} ^ {n}b_{i}\),，则当 \(n \geq 2\) 时，\(b_{i} = B_{i} - B_{i - 1}\)，且 \(B_{1} = b_{1}\) ，于是当 \(n \geq 2\) 时：

\[\begin{aligned} \sum\limits_{i = 1} ^ {n} c_ib_i &= \sum\limits_{i = 2} ^ {n} c_{i} (B_{i} - B_{i - 1}) + c_{1}b_{1} \\ &= \sum\limits_{i = 2} ^ {n} c_{i}B_{i} - \sum\limits_{i = 1} ^ {n - 1} c_{i + 1}B_{i} + c_{1}b_{1} \\ &= \sum\limits_{i = 1} ^ {n} c_{i}B_{i} - \sum\limits_{i = 1} ^ {n - 1} c_{i + 1}B_{i}\\ &= \sum\limits_{i = 1} ^ {n - 1} (c_{i} - c_{i + 1})B_{i} + c_nB_{n} \end{aligned} \]

等差 \(\times\) 等比型数列求和

可以把等差数列推广到任意多项式数列的求和）

如果 \(c_{n}\) 是等差数列，那么 \(c_{n + 1} - c_{n}\) 就是一个常数， \(b_{n}\) 是等比数列，那么 \(B_{n}\) 可以通过等比数列求和快速计算，这样就能计算出 \(\sum\limits_{i = 1} ^ {n}c_{i}b_{i}\).

例：\(c_{n} = n, \ b_{n} = \frac{1}{2^{n}}\) ，求 \(T_{n} = \sum\limits_{i = 1} ^ {n} c_{i}b_{i}\).

由 Abel 变换：

\[\begin{aligned} T_{n} &= \sum\limits_{i = 1} ^ {n} c_{i}b_{i}\\ &= \sum\limits_{i = 1} ^ {n - 1}(-1)(1 - \frac{1}{2 ^{n}}) + n \times (1 - \frac{1}{2^{n}}) \\ &=1 - \frac{1}{2^{n - 1}} - n + 1 + n - \frac{n}{2^{n}} \\ &= 2 - \frac{n + 2}{2 ^ {n}} \end{aligned} \]

例：求 \((2n - 1)2^{n}\) 的前 \(n\) 项和.

\[\begin{aligned} T_{n} &= \sum\limits_{i = 1} ^ {n} (2n - 1)2^{n} \\ &= \sum\limits_{i = 1} ^ {n - 1}-2 \times (2 ^ {i + 1} - 2) + (2n - 1) \times (2 ^ {n + 1} - 2) \\ &= (2n - 3) 2^{n + 1} + 6 \end{aligned} \]

例： 求 \(n ^ {2}\) 的前 \(n\) 项和.

设 \(c_n = b_n = n\) ，那么由 Abel 变换：

\[\begin{aligned} T_{n} &= \sum\limits_{i = 1} ^ {n} n \times n \\ &= \sum\limits_{i = 1} ^ {n - 1} -1 \times \frac{i(i+1)}{2} + \frac{n^{2}(n +1)}{2} \\ &= -\frac{1}{2} \times (T_{n - 1} + n ^ {2} + \frac{n(n - 1)}{2} - n^{2}) + \frac{n^{2}(n + 1)}{2} \\ \frac{3}{2} T_{n} &= -\frac{1}{2}(\frac{-n^{2} - n}{2}) + \frac{n^{2}(n + 1)}{2}\\ &= \frac{n(n + 1)(2n + 1)}{4} \\ T_{n} &= \frac{n(n+1)(2n + 1)}{6} \end{aligned} \]

按这个思路，我们可以求出数列 \({n^{m}}\) 的前 \(k\) 项和。