2024.11.2
Part1 基础部分
【例1】已知 $y = |2x+6| + |x-1| - 4|x+1|$,求 $y$ 的最大值。
解:直接零点分段即可,易得 $y$ 的最大值为 $6$。
Part2 $$ \mathscr{Abel} $$变换
【例1】若已经给出两个数列 $ {a_n} , {b_n} $,构造数列 $ {c_n} $ 使得 $ c_n = a_nb_n $,令 $T_n = \sum\limits_{i = 1}^n c_i$,$B_n = \sum\limits_{i = 1}^{n} b_i$。
求证:$T_n = (a_1 - a_2)B_1 + (a_2 - a_3)B_2 + \dots + (a_{n-1} - a_n)B_{n - 1} + a_nB_n = \sum\limits_{i = 1}^{n - 1} (a_i - a_{i + 1})B_i + a_nB_n$。
$proof:$
$$ T_n = a_1b_1 + a_2b_2 + \dots + a_nb_n$$
$$ = a_1B_1 + a_2(B_2 - B_1) + \dots + a_n(B_n - B_{n - 1}) $$
$$ = (a_1 - a_2)B_1 + (a_2 - a_3)B_2 + \dots + (a_{n-1} - a_n)B_{n - 1} + a_nB_n $$
【例2】求证 $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \dfrac{1}{6}n(n+1)(2n+1)$。
$proof:$ 利用 Abel 变换可以构造,$c_n = n \times n = a_n \times b_n,a_n = b_n = n$,此时 $B_n = \frac{n^2 + n}{2}$ 则
$$ T_n = c_1 + c_2 + \dots + c_n $$
$$ = 1^2 + 2^2 + 3^2 + \dots + n^2 $$
$$ = (a_1 - a_2)B_1 + (a_2 - a_3)B_2 + \dots + (a_{n-1} - a_n)B_{n - 1} + a_nB_n $$
$$ = -B_1 - B_2 - B_3 - \dots - B_{n - 1} + nB_n $$
$$ = -(\frac{1^2 + 1}{2} + \frac{2^2 + 2}{2} + \dots + \frac{(n - 1)^2 + (n - 1)}{2}) + nB_n $$
$$ = -\frac{1}{2} \times ((1^2 + 2^2 + \dots + (n - 1)^2) + (1 + 2 + \dots + (n - 1))) + \frac{n^3 + n^2}{2} $$
$$ = -\frac{1}{2} \times (T_n - n^2 + \frac{n^2 - n}{2}) + \frac{n^3 + n^2}{2} $$
$$ = -\frac{1}{2}T_n + \frac{1}{2}n^2 - \frac{n^2 - n}{4} + \frac{n^3 + n^2}{2} $$
$$ = -\frac{1}{2}T_n + \frac{2n^3 + 3n^2 + n}{4} $$
$$ \therefore \frac{3}{2}T_n = \frac{n(n + 1)(2n + 1)}{4} $$
$$ T_n = \frac{1}{6}n(n+1)(2n+1) $$
标签:dots,frac,limits,第一,nB,times,绝对值,有理数,2n From: https://www.cnblogs.com/LiCX1/p/18523410