update some errors

thu_dsa/chp1.md

@@ -151,7 +151,7 @@ _若存在函数f(n)和正的常数c1和c2，使得在渐进的条件(n >> 2)下
 + 幂方级数: 比幂次高出一阶
 	- $T_2(n) = 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} = O(n^3)$
 	- $T_3(n) = 1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^{2}(n+1)^{2}}{4} = O(n^4)$
-	- $T_4(n) = 1^4 + 2^4 + 3^4 + ... + n^4 = \frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30} = O(n^3)$
+	- $T_4(n) = 1^4 + 2^4 + 3^4 + ... + n^4 = \frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30} = O(n^5)$
 	- ...
 	- $T_k(n) = 1^k + 2^k + 3^k + ... + n^k = \int_{0}^{n}x^k{d}x = \frac{x^{k+1}}{k+1}|_{0}^{n} = O(n^{k+1}))$
 + 几何级数: 与末项同阶