## 4 thoughts on “Tổng và tích (1)”

1. trungtuan says:

Cach tiep can dep de nhat doi voi cac tong huu han hay cac tich huu han theo toi do la su dung dinh li sau:

Dinh li. Cho day cac so $a_1,a_2,...,a_n,a_{n+1}$. Khi do
a)$\sum_{i=1}^n(a_{i+1}-a_i)=a_{n+1}-a_1$.
b) Neu co them tat ca cac so hang cua day khac khong thi $\prod_{i=1}^n\dfrac{a_{i+1}}{a_i}=\dfrac{a_{n+1}}{a_1}$.

Chung minh. De! $\Box$

2. Nguyen Ngoc Quy says:

mot vai bai tap ap dung
Bai 1.
Tinh $\sin 1 + \sin 2 + ... + \sin n$.
Bai 2.
Tinh $\lim_{n\to\infty}\prod_{i=2}^n\dfrac{i^3-1}{i^3+1}$.
Bai 3.
Tính $\dfrac{1}{1+\cot 1^0}+\dfrac{1}{1+\cot 5^0}+...+\dfrac{1}{1+\cot 89^0}$.
Bai 4.
Tinh $\tan^2{1^0}+\tan^2{3^0}+...+\tan^2 {89^0}$.

3. trungtuan says:

Bai 5.([1])
Tinh $\sum_{k=1}^n\cos kx$.
Bai 6.([1])
Tinh $\sum_{k=0}^n\tan^{-1}\dfrac{1}{k^2+k+1}$.
Bai 7.([1])
Chung minh rang $\sum_{i=1}^n\dfrac{\sin ix}{\cos^i x}=\cot x-\dfrac{\cos (n+1)x}{\sin x\cos^n x}$ voi moi $x\not = \dfrac{k\pi}{2},k\in\mathbb{Z}$.
Bai 8.([1])
Chung minh rang $\sum_{i=0}^{88}\dfrac{1}{\cos i^{0}\cos (i+1)^0}=\dfrac{\cos 1^{0}}{\sin^2 1^{0}}$.

Tai lieu tham khao:
[1] Mathematical Olympiad Challenges, Titu Andreescu and Razvan Gelca, Birkhauser.