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NL
7
9 tháng 5 2017
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+.....+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{8}{9}+\frac{1}{2}\cdot\frac{2}{5}=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
9 tháng 5 2017
=1/2x(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.....+1/8-1/10)
=1/2x58/45
=29/45
MV
29 tháng 4 2017
\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98
C
0
HX
2
BM
24 tháng 4 2019
mk ko ghi đb nhé
\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)
\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)
\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)
\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
phần còn lại bn tự tính nha
4 tháng 4 2019
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
NT
1
10 tháng 9 2016
B=1.3+2.4+3.5+...+97.99+98.100
B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)
B=1.2+1+2.3+2+....+97.98+97+98.99+98
B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)
B=98.99.100/3 + 98.99/2
B=323400+4851=328251
uh phep cuoi la 98.100 nhaa
hình như bn chép sai đề
hình như sai đề thì phải
Ta có:
\(A=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+98\left(98+2\right)\)
\(A=\left(1\cdot1+1\cdot2\right)+\left(2\cdot2+2\cdot2\right)+\left(3\cdot3+3\cdot2\right)+\cdots+\left(98\cdot98+98\cdot2\right)\)
\(A=\left(1^2+2^2+3^2+\cdots+98^2\right)+2\left(1+2+3+\cdots+98\right)\)
Đặt \(B=1^2+2^2+3^2+\cdots+98^2\)
và \(C=1+2+3+\cdots+98\)
Ta có: \(B=1^2+2^2+3^2+\cdots+98^2\)
\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+\cdots+98\left(99-1\right)\)
\(B=\left(1\cdot2+2\cdot3+3\cdot4+\cdots+98\cdot99\right)-\left(1+2+3+\cdots+98\right)\)
\(\Rightarrow A=\left(1\cdot2+2\cdot3+3\cdot4+\cdots+98\cdot99\right)+\left(1+2+3+...+98\right)\)
Dặt \(D=1\cdot2+2\cdot3+3\cdot4+\cdots+98\cdot99\), tính tổng \(D\) :
\(D=1\cdot2+2\cdot3+3\cdot4+\cdots+98\cdot99\)
\(3D=1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+3\cdot4\left(5-2\right)+\cdots+98\cdot99\left(100-97\right)\)
\(3D=\left(1\cdot2\cdot3-0\cdot1\cdot2\right)+\left(2\cdot3\cdot4-1\cdot2\cdot3\right)+\left(3\cdot4\cdot5-2\cdot3\cdot4\right)+\cdots+\left(98\cdot99\cdot100-97\cdot98\cdot99\right)\)
\(3D=98\cdot99\cdot100-0\cdot1\cdot2\)
\(3D=98\cdot99\cdot100\)
\(D=98\cdot33\cdot100\)
\(C=1+2+3+\cdots+98\)
\(C=\left(98+1\right)\cdot\frac{98}{2}\)
\(C=4851\)
Vậy \(A=D+C=98\cdot33\cdot100+4851=328251\)
Tích cho mik nha
sao khó v tr
Sửa đề: \(1\cdot3+2\cdot4+3\cdot5+\cdots+98\cdot100\)
Đặt \(A=1\cdot3+2\cdot4+\cdots+98\cdot100\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+98\left(98+2\right)\)
\(=\left(1^2+2^2+\cdots+98^2\right)+2\left(1+2+\cdots+98\right)\)
\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)
\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=33\cdot49\cdot197+98\cdot99\)
=318549+9702
=328251