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a ) Co :
1/1.2 - 1/2.3 = 2/1.2.3
1/2.3 - 1/3.4 = 2/2.3.4
...
1/37.38 - 1/38.39 = 2/37.38.39
=> 2M = 2/1.2.3 + 2/2.3.4 + ... + 2/37.38.39
=> 2M = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/37.38 - 1/38.39
=> 2M = 1/2 - 1/1482
=> 2M = 370/741
=> M = 185/741
B ) A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8
3A = 1 + 1/3 + 1/3^2 + ... + 1/3^7
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^7 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8 )
2A = 1 - 1/3^8
A = ( 1 - 1/3^8 ) / 2
a/ \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+........+\frac{99}{100!}\)
\(\Leftrightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+......+\frac{100-1}{100!}\)
\(\Leftrightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+.....+\frac{100}{100!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{99!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{100!}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}\)
1)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Sửa đề chút nha
\(\frac{x}{2}=\frac{1}{1.2.3}+....+\frac{1}{98.99.100}\)
Ta có công thức tổng quát \(\frac{1}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{2}\left(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right)\)
\(\Rightarrow\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)
Áp dụng vào tổng ta có
\(\frac{x}{2}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow x=\frac{4949}{4950}\)
Mình nghĩ bài 1 là rút gọn biểu thức nên sẽ giải như này:
Bài 1
\(B=3+3^2+3^3+...+3^{2015}\)(1)
Nhân 2 vế của (1) với 3
3B= \(3^2+3^3+3^4...+3^{2016}\)(2)
Trừ 2 vế của (2) cho (1)
3B-B= \(\left(3^2+3^3+3^4...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)
2B =\(3^2+3^3+3^4...+3^{2016}-3-3^2-3^3-...-3^{2015}\)
2B =\(\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2015}-3^{2015}\right)+\left(3^{2016}-3\right)\)
2B =\(3^{2016}-3\)
B = \(\frac{\left(3^{2016}-3\right)}{2}\)
Bài 2 làm tương tự như số mũ sẽ giảm đi
nhưng phần tìm n thì mình ko biết
Bài 3
nhân 2 vế với \(\frac{1}{2}\)ta có 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100=1/1.2-1/99.100
=>1/1.2-1/99.100=1/k.(1/1.2-1/99.100)
=>1/k=1=>k=1
Bài 4:
rút gọn lại dc 5/28+5/70+5/130+...+5/700
tách 28 thành 4.7; 70 thành 7.10; 130 thành 10.13 ...
nhân cả biểu thức với 5/3
5/3A= 1/4-1/7+1/7-1/10+1/10-1/13+...+1/25-1/28
5/3A= 1/4-1/28
5/3A= 3/14
A=9/70
Bài 5: Vì 1/2<2/3;3/4<4/5;5/6<6/7...99/100<100/101
=>M<N
\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)
\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)
\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
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
a) \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{9900}\)
\(A=\frac{9898}{19800}.\)
Vậy :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{9898}{19800}:2\)
\(A=\frac{4949}{19800}.\)
a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
A = \(\frac{1}{2}.\frac{4949}{9900}\)
A = \(\frac{4949}{19800}\)