A = 1.2 + 2.3 + 3.4 + ....+ 101.102

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 101.102.3

=> 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + .... + 101.102.(103-100)

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 101.102.103 - 100.101.102

=> 3A = 101.102.103

=> A = \(\frac{101\cdot102\cdot103}{3}=353\text{ }702\)

A = 1 . 2 + 2 . 3 + 3 . 4 + ...+ 101 . 102 

3A = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ...+ 101 . 102 . 3 

3A = 1 . 2 . 3 + 2. 3 . ( 4 - 1 ) + 3. 4 . ( 5 - 2 ) + ...+ 101 . 102 . ( 103 - 100 ) 

3A = 1 . 2 . 3 + 2. 3 . 4 - 2 . 3 . 1 + 3 . 4 . 5 - 2. 3 . 4 + ...+ 101 . 102 . 103 - 101. 102 . 100 

3A = (  1. 2 . 3 - 1. 2. 3 ) + ( 2 . 3 . 4 - 2. 3 . 4 ) + ...+ ( 99 . 100 . 101 - 99 . 100 . 101 ) + 101 . 102 . 103 

3A = 101 . 102 . 103 

  A = ( 101 . 102 . 103 ) : 3 

 A  = 353702 

Tham khảo nha !!! 

số học sinh giỏi của lớp 6A là:

48.1/3=16(hs)

còn lại:

48-16=32(hs)

số hs khá:

32.3/4=24(hs)

số hs trung bình:

48-(16+24)=8(hs)

k mk nha. kb nữa

3A=3.1.2+3.2.3+....+3.101.102

=1.2(3-0)+2.3(4-1)+....+102.102(103-100)

=1.2.3-0.1.2+2.3.4-1.2.3+....+101.102.103-100.101.102

=-0.1.2+101.102.103

=101.102.103=1061106

a) Gọi d là ƯCLN (n+1;2n+5)

\(\Rightarrow\hept{\begin{cases}n+1⋮d\\2n+5⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+5⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+5⋮d\end{cases}}}\)

\(\Rightarrow\left(2n+5\right)-\left(2n+2\right)⋮d\)

\(\Rightarrow3⋮d\Rightarrow d\in\left\{1;3\right\}\)

Mà 2n+2 ko chia hết cho 3

=>d=1

Vậy......

b)Gọi d là ƯCLN(2n+3;2n+8)

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}}\)

\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)

\(\Rightarrow2⋮d\Rightarrow d\in\left\{1;2\right\}\)

Mà 2n+3 ko chia hết cho 2

\(\Rightarrow d=1\)

Vậy.......