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5 tháng 6 2017

Đặt A = 1 x 2 + 2 x 3 + 3 x 4 + ... + n x ( n - 1)
=> 3A = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + n x (n - 1) x [(n + 2) x (n + 1)]
=> 3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + ... + n x (n + 1) x (n + 2)
=> 3A = n x (n + 1) x (n + 2)
=> A = n x (n + 1) x (n + 2) / 3

5 tháng 6 2017

3S=1.2.3+3.4.5+...+n.(n-1).3

1.2.(3-0).......................................................

 k mk đi mk giải tiếp cho nha

29 tháng 8 2016

Đặt A = 1x2+2x3+3x4+...+nx(n+1)

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + n.(n + 1).[(n + 2).(n - 1)]

=> 3A =  1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n + 1).(n + 2)

=> 3A = n.(n + 1).(n + 2)

=> A = n.(n + 1).(n + 2) / 3

5 tháng 6 2017

Cách làm mk làm giống  Edokawa Conan nhé kw ;\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

3 tháng 8 2017

cac cau tu di ma lam bai tap cua minh

3 tháng 8 2017

cau len mang di , bai nay mk chua hoc , sory nha

chuc ban hoc tot ^-^

4 tháng 2 2016

kết quả là 6015

Đặt \(A=1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103\)

\(=1\cdot2\cdot\left(3+1\right)+2\cdot3\cdot\left(4+1\right)+\cdots+100\cdot101\cdot\left(102+1\right)\)

\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

Đặt \(B=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\)

\(=\left(2-1\right)\cdot2\cdot\left(2+1\right)+\left(3-1\right)\cdot3\cdot\left(3+1\right)+\cdots+\left(101-1\right)\cdot101\cdot\left(101+1\right)\)

\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+101\left(101^2-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+3^3+\cdots+101^3\right)-1-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1+2+3+\cdots+101\right)\)

\(=\left(1+2+3+\cdots+101\right)^2-\left(1+2+3+\cdots+101\right)\)

\(=\left\lbrack101\cdot\frac{102}{2}\right\rbrack^2-101\cdot\frac{102}{2}=\left(101\cdot51\right)^2-101\cdot51\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+100\cdot101\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+100\left(100+1\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{100\cdot101}{2}=\frac{100\cdot101\cdot201}{6}+50\cdot101\)

\(=50\cdot101\cdot67+50\cdot101=50\cdot101\cdot68\)

Ta có: A\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

=B+C

\(=\left(101\cdot51\right)^2-101\cdot51+50\cdot101\cdot68\)

\(=101^2\cdot51^2-101\cdot51+50\cdot101\cdot68=101\left(101\cdot51^2-51+50\cdot68\right)=101\cdot266050\)

Đặt \(D=1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+101^2\left(101-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2^2+3^2+\cdots+101^2\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1^2+2^2+\cdots+101^2\right)\)
\(=\left(1+2+\cdots+101\right)^2-101\cdot\frac{\left(101+1\right)\left(2\cdot101+1\right)}{6}\)

\(=\left(101\cdot\frac{102}{2}\right)^2-101\cdot17\cdot2023=101^2\cdot51^2-101\cdot17\cdot2023\)

\(=101\cdot17\left(101\cdot17\cdot3^2-2023\right)=101\cdot17\cdot13430\)

Ta có: \(\frac{1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103}{1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2}\)

\(=\frac{101\cdot266050}{101\cdot17\cdot13430}=\frac{1565}{1343}\)

10 tháng 8 2017

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{n\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(=\frac{n+1}{n+1}-\frac{1}{n+1}\)

\(=\frac{n}{n+1}\)

19 tháng 2 2016

vì 1/1*2=1-1/2

   1/2*3=1/2-1/3

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

1/2014*2015=1/2014-1/2015

=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015

=1-1/2015

=2014/2115

19 tháng 2 2016

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

9 tháng 11 2020

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{1}-\frac{1}{6}\)

\(=\frac{5}{6}\)

13 tháng 7 2021

\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)

\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)

\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)

\(=\)\(\frac{5}{6}\)

Hok tốt

\(C=-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-....-\frac{1}{42}+\frac{1}{43}-\frac{1}{43}+\frac{1}{44}\)

\(C=-1+\frac{1}{44}\)

\(C=-\frac{43}{44}\)

8 tháng 4 2016

C= \(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{43.44}\right)=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{43}-\frac{1}{44}\right)\)

 =\(-\left(1-\frac{1}{44}\right)=-\frac{43}{44}\)

12 tháng 8 2016

Đặt A , ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(A=2-\frac{1}{1000}\)

\(A=\frac{2000}{1000}-\frac{1}{1000}\)

\(A=\frac{1999}{1000}\)

12 tháng 8 2016

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}+1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+1\)

\(A=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

Vậy \(A=\frac{1999}{1000}\)