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\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
\(S=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{24.25}\right)\)
\(S=\frac{1}{4}-\frac{1}{24.50}\)
Dễ thấy với mọi số tự nhiên n > 1 , ta có :
\(\frac{2}{\left(n-1\right).n.\left(n+1\right)}=\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}\)
Sử dụng hệ thức trên cho từng số hạng trong tổng sau :
\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(n-1\right).n.\left(n+1\right)}+\frac{2}{23.24.25}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}+...+\frac{1}{23.24}-\frac{1}{24.25}\)
Để ý rằng trong vế phải của hệ thức trên , trừ 2 số hạng đầu và cuối , các số hạng còn lại tạo thành từng cặp đối nhau.
Do đó , có thể rút gọn :
\(2S=\frac{1}{1.2}-\frac{2}{24.25}=\frac{299}{600}\)
Vậy , ta được \(S=\frac{299}{600}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right).n.\left(n+1\right)}+...+\frac{1}{23.24.25}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{24.25}\right)=\frac{299}{1200}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{600}\right)=\frac{1}{2}.\frac{299}{600}=\frac{299}{1200}\)
A=1/1.2.3 + 1/2.3.4 + ... + 1/23.24.25
2A=2/1.2.3 + 2/2.3.4 + ... + 2/23.24.25
=1/1.2 - 1/2.3 + 1/2.3 -1/3.4 + .... + 1/23.24 - 1/24.25
=1/1.2 - 1/24.25
Tớ chỉ giải đến đó thôi còn lại các bạn cứ bấm máy tính là ra
Bài toán trên áp dụng bài toán tổng quát sau:
\(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}=\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
Suy ra
\(\frac{1}{n.\left(n+1\right).\left(n+2\right)}=\left(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\right).\frac{1}{2}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right).\frac{1}{2}\)
\(=\left(\frac{1}{1.2}-\frac{1}{24.25}\right).\frac{1}{2}\)
\(=\frac{299}{1200}\)
3/1*3+3/3*5+......+3/99*101
=3/2*(2/1*3+3/3*5+.............+2/99*101)
=3/2*(1-1/3+1/3-1/5+..........+1/99-1/101)
=3/2*(1-1/101)
=3/2*100/101
=150/101
Câu 1
=>S=2/3( 2/(1.3) + 2/(3.5)+.....+ 2/(99.101) )
=>S=2/3(1-1/3+1/3-1/5+...+1/99-1/101)
=>S=2/3(1-1/101)
=>S=2/3.100/101
=>S=200/303
2/ \(\frac{2}{3}S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{25-23}{23.24.25}\)
\(\frac{2}{3}S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}=\frac{1}{2}-\frac{1}{24.25}\Rightarrow S=\left(\frac{1}{2}-\frac{1}{24.25}\right):\frac{2}{3}\)
1/
\(\frac{2}{3}S=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\)
\(\frac{2}{3}S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\Rightarrow S=\frac{100}{101}.\frac{3}{2}=\frac{150}{101}\)
Ta có nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
B=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
Vậy \(B=\frac{185}{741}\)
Ta có nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.29}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{185}{741}\)
Tính:
S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2S=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4949}{9900}\)
\(S=\frac{4949}{19800}\)
Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
...
\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)
=> 2S = \(\frac{4949}{9900}\)
=> S = \(\frac{4949}{19800}\)