Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)
Với n là số tự nhiên khác 0; Chứng minh \(\dfrac{1\cdot3\cdot5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...\left(n+n\right)}=\dfrac{1}{2^n}\)
Help me please!
chứng tỏ rằng với mọi n thuộc N ta luôn có
\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
Cho n + 1 tia chung gốc. Hỏi tạo được bao nhiêu góc?
\(\dfrac{\left(n+1\right)n}{2}\)
\(\dfrac{\left(n-1\right)n}{2}\)
(n +1 ).n
\(\dfrac{n^2}{2}\)
Tìm n biết:
a) \(\dfrac{32}{\left(-2\right)^n}=4\)
b) \(\dfrac{8}{2^n}\)\(=2\)
c) \(\left(\dfrac{1}{2}\right)^{2n-1}\)\(=\dfrac{1}{8}\)
chứng minh rằng \(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
Tìm tất cả các số nguyên tố p có dạng \(\dfrac{n\left(n+1\right)}{2}-1\left(n\ge1\right)\)
Tìm số nguyên n thoả mãn:\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{2022}{2023}\) SOS giúp tôi với
\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
