Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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:
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.29}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
2B = 2.\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\right)\)
<=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{37.38.39}\)
2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
2B = \(\frac{1}{1.2}-\frac{1}{38.39}\)
B = \(\left(\frac{1}{2}-\frac{1}{38.39}\right):2\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}=\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(=\left(\frac{1}{2}-\frac{1}{38\cdot39}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)+...+\left(\frac{1}{37\cdot38}-\frac{1}{37\cdot38}\right)=\left(\frac{741}{1482}-\frac{1}{1482}\right)+0+...+0=\frac{740}{1482}=\frac{370}{741}\)Chúc bạn học tốt!^_^
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...........+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..........+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{1}{2}-\frac{1}{1482}\)
\(=\frac{741}{1482}-\frac{1}{1482}\)
\(=\frac{370}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\frac{740}{1482}\)
\(=\frac{185}{741}\)
Chúc bạn học tốt !!!
Đặt 1/1.2.3 + 1/2.3.4 + ...+ 1/37.38.39 = A
Ta có : 2A = 2/1.2.3 + 2/2.3.4 +...+ 2/37.38.39
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/37.38 - 1/38.39
2A = 1/1.2 - 1/38.39
2A = 740/1482 = 370/741
A= 370/741 . 1/2 =........
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\frac{370}{741}\)
\(=\frac{185}{741}\)
Tổng quát : \(\frac{2}{(a-1)\cdot a\cdot(a+1)}=\frac{1}{a(a-1)}-\frac{1}{a(a+1)}\)
Ta có : \(2A=2\cdot(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{37\cdot38\cdot39})\cdot1428+185\cdot8\)
\(\Leftrightarrow2A=(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{37\cdot38\cdot39})\cdot1428+185\cdot8\)
\(\Leftrightarrow2A=[(\frac{1}{1\cdot2}-\frac{1}{2\cdot3})+(\frac{1}{2\cdot3}-\frac{1}{3\cdot4})+...+(\frac{1}{37\cdot38}+\frac{1}{38\cdot39})]\cdot1428+185\cdot8\)
\(\Leftrightarrow2A=(\frac{1}{1\cdot2}-\frac{1}{38\cdot39})\cdot1428+185\cdot8=\frac{541680}{247}\)
\(\Leftrightarrow A=\frac{541680}{494}=1096,518219\approx1096,5\)
Chúc bạn học tốt
A = ( 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/37.38.39 ) . 1428 + 185 . 8
A = 1/2( 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/37.38.39 ) . 1428 + 185.8
A = 1/2( 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/37.38 - 1/38.39 ) . 1428 + 185.8
A = 1/2( 1/1.2 - 1/38.39 ) . 1428 + 185.8
A = 1/2 . 370/741 . 1428 + 185.8
A = 185/741 . 1428 + 185 . 8
A = 185 . 1428 : 741 + 185 . 8
A = 185 . 476/247 + 185 . 8
A = 185 ( 476 / 247 + 8 )
A = 185. 2452/247
A = 1836,518219
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{185}{741}\)
3765942