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C=\(\frac{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}{4}\)
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
\(M=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1482+185.8\)
a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0
=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0
b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4
=1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)
= (x-1)x(x+1)(x+2)
=> A=x(x+1)(x+2)(x-1)/4
Mình không chép đề bài nhé :
Gọi biểu thức là A :
Ta có : 2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
=\(\frac{1}{1.2}-\frac{1}{49.50}\)( Rút gọn đi ta được cái này )
=1/2 - 1/2450
Vậy A = (1/2 - 1/2450):2
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
= \(\frac{1}{1.2}-\frac{1}{49.50}\)
= \(\frac{1}{2}-\frac{1}{2450}\)
= \(\frac{612}{1225}\)
đặt
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1.2}-\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2450}=\frac{621}{1225}\)
\(\Rightarrow A=\frac{306}{1225}\)
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
\(2A=2\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\right).1428+1480\)
\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right)\times1428+1480\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\times1428+1480\)
\(=\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\times1428+1480\)
\(=\left(\frac{741}{1482}-\frac{1}{1482}\right)\times1428+1480\)
\(=\frac{740}{1482}\times1428+1480\)
\(=\frac{528360}{741}+1480\)
Vongola: Em tách đúng tuy nhiên A còn hạng tử đằng sau, em nhân 2 thì phải nhân cả hạng tử đó nữa. Tức là không phải 1480 mà là 2960 e nhé :)
A = 1.2.3 + 2.3.4 + ....+ 48.49.50
=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)
= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50
= 48.49.50.51
=> A = 48.49.50.51:4 = 12.49.50.51
bài b) làm tương tự nha