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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 \(S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\frac{6533}{39200}=\frac{6533}{117600}\)
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:
\(S=\frac{1}{3}\left[\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right].\)
Giải tiếp(ko chép đề)
= 1/1 - 1/2 - 1/3 - 1/4 + 1/2 - 1/3 - 1/4 - 1/5 + ... + 1/27 - 1/28 - 1/29 - 1/30
= 1 - 1/30
= 29/30
ks nha
Bài giải :(không chép đề)
=1-1/2-1/3-1/4-1/5+1/2-1/3-1/4-1/5+........+1/27-1/28-1/29-1/30
=1-1/30
=29/30
Vậy số cần tìm là:29/30 Suy ra Y=29/30
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{970200}\right)=\frac{1}{18}-\frac{1}{6.970200}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}\)
\(=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+ \frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\frac{161699}{970200}=\frac{161699}{299106000}\)
\(S=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{20.21.22.23}\)
\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+......+\frac{3}{20.21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.....+\frac{1}{20.21.22}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}.\frac{295}{1771}=\frac{295}{3542}\)