K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 4 2016

a)Ta có:\(A=2^2+4^2+...+20^2\)

\(\Rightarrow A=2^2\left(1^2+2^2+...+10^2\right)\)

\(\Rightarrow A=4.204=816\)

Vậy A=816

29 tháng 4 2016

khó quá à bạn ơi

22 tháng 7 2019

#)Giải :

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}=7\left[\frac{1}{7}\left(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\right)\right]\)

\(C=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\times\frac{3}{35}=\frac{21}{35}\)

22 tháng 7 2019

#)Góp ý :

Ý D tương tự ý C

I don't now

mik ko biết 

sorry 

......................

25 tháng 7 2018

b,\(B=2^2+4^2+...+20^2\)

\(\Rightarrow B=2^2\left(1^2+2^2+...+10^2\right)\)

\(\Rightarrow B=4.\left[1.\left(2-1\right)+2.\left(3-1\right)+...+10.\left(11-1\right)\right]\)

\(\Rightarrow B=4\left(1.2-1+2.3-2+...+10.11-10\right)\)

\(\Rightarrow B=4\left[\left(1.2+2.3+...+10.11\right)-\left(1+2+...+10\right)\right]\)

\(\Rightarrow B=4\left(\frac{10.11.12}{3}-\frac{11.10}{2}\right)\)

1) A=7/10.11+7/11.12+7/12.13+...+7/69.70

  A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)

 A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)

 A= 7.(1/10-1/70)

 A=7.3/35=3/5

2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75

  B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

  B=1/25-1/75=2/75

19 tháng 6 2016

A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70

(1/7).A=1/10.11+1/11.12+...+1/69.70

=1/10-1/11+1/11-1/12+...+1/69-1/70

=1/10-1/70=3/35

=>A=7.(3/35)

=3/5

2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75

=>(1/2).B=2/25.27+...+2.73.75

=1/25-1/27+...+1/73-1/75

=1/25-1/75

=2/75

=>B=4/75

7 tháng 3 2018

\(B=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{19.20}\)

\(B=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{19.20}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{20}\right)\)

\(B=5.\frac{1}{20}=\frac{1}{4}\)

7 tháng 3 2018

\(C=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\)

\(4C=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(4C=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\)

\(4C=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)

\(4C=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)

\(C=\frac{4}{25}:4=\frac{1}{25}\)

4 tháng 10 2015

3N = 1.2.3+2.3(4-1)+3.4.(5-2)+.+99.100.(101-98)

3N = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.+99.100.101-98.99.100

3N = 99.100.101

3N=33.100.101=333300

b)

tổng này có  99-10+1=90 (số hạng):

10,11 + 11,12 + 12,13 +............+ 98,99 + 99,100 =

10,100 + 11,11 + 12,12 + .......... + 98,98 + 99,99 =

(10,10 + 99,99) x 90 : 2 = 4954,05

c)

R=1.(2-1)+2.(3-1)+.....+100.(101-1)

=1.2-1.1+2.3-1.2+......+100.101-1.100

=(1.2+2.3+.....+99.100+100.101)-(1+2+3+...+100)

=[1.2.3+2.3.(4-1)+........100.101.(102-99)]:3+[(100+1).100:2]

(tổng trên chia cho 3 nên cuối cùng chia 3)

=(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....100.101.102-99.100.101):3+5050

=(100.101.102) :3 +5050

=348450

d)=1.100+2.(100-1)+.....+100.(100-99)

=1.100+2.100-1.2+3.100-2.3+........+100.100-99.100

=100.(1+2+3+.......+100)-(1.2+2.3+3.4+....+99.100)

=100.\(\frac{101.100}{2}-\frac{99.100.101}{3}\) =505000-333300=171700

p/s mỏi tay, bấm mình nhé

 

      

b) Ta có: \(B=\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}+\dfrac{1}{13\cdot14}\)

\(=\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=\dfrac{1}{10}-\dfrac{1}{14}\)

\(=\dfrac{14}{140}-\dfrac{10}{140}\)

\(=\dfrac{4}{140}=\dfrac{1}{35}\)