Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét vế phải: \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
= \(\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
= \(\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}\)
= \(\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
= VT
=> Đpcm
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm
Biến đổi ở phân số dạng tổng quát :
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{3}{3n(n+1)(n+2)(n+3)}=\frac{3+n-n}{3n(n+1)(n+2)(n+3)}\)
\(=\frac{1}{3}\left[\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Áp dụng kết quả này vào bài được :
\(\frac{1}{1\cdot2\cdot3\cdot4}=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}\right],\frac{1}{2\cdot3\cdot4\cdot5}=\frac{1}{3}\left[\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}\right],...\)
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Cộng từng vế,ta được : \(S=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
P/S : Xong
Ta có: S= \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3S=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(\Rightarrow S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
Vậy \(S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
Ta có:
\(1.3.5.7.9...\left(2n-1\right)=\frac{\left[1.3.5.7.9....\left(2n-1\right)\right].\left[2.4.6.8...2n\right]}{2.4.6.8....2n}=\frac{1.2.3.4.5.6....2n}{\left(2.1\right).\left(2.2\right).\left(2.3\right)\left(2.4\right)....\left(2.n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{1.2.3.4.5.6....2n}{\left(2.2.2.....2\right).\left(1.2.3.4.....n\right)}=\frac{\left(1.2.3.4.....n\right)\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}{2^n.\left(1.2.3.4....n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n}\)
=> \(\frac{1.3.5.7.9...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}=\frac{1}{2^n}\)(đpcm)
Biến đổi phân số ở dạng tổng quát:
\(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{3}{3n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{3+n-n}{3n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{3}\left[\frac{n+3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)\left(n+2\right)}\right]\)
=\(\frac{1}{3}\left[\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right]\)
Áp dụng kết quả vào bài, ta được:
\(\frac{1}{1.2.3.4}=\frac{1}{3}\left[\frac{1}{1.2.3}-\frac{1}{2.3.4}\right],\frac{1}{2.3.4.5}=\frac{1}{3}\left[\frac{1}{2.3.4}-\frac{1}{3.4.5}\right]\),...
\(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{1}{3}\left[\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right]\)
Cộng từng vế, ta được:
\(S=\frac{1}{3}\left[\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right].\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
