Ta có : abba = 1001a + 110b 

Mà 1001 chai hết cho 11 và 110 chai hết cho 11

Nên 1001a chia hết cho 11 và 110b chia hết cho11

Suy ra abba chia hết cho 11

Ta có: S = 1.2 + 2.3 + 3.4 + ....... + 99.100 + 100.101

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ....... + 100.101.102

=> 3S = 100.101.102

=> S = 100.101.102 / 3

=> S = 343400

các bạn cố làm hết giúp mik nha !!!

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

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

(6x + 35 ) = 330 : 6

6x+35=55

6x=55-35

6x=20

x=20:6

x=10/3

có công thức nè bạn 1²+2²+3³+...+n²= n(n+1)(2n+1):6

\(\frac{B}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\)

\(\frac{B}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)

\(\frac{B}{2}=\frac{100}{101}\)

\(B=\frac{200}{101}\)

B = \(2\left(\frac{1}{1x2}+\frac{1}{2x3}+....+\frac{1}{100x101}\right)\)

B = \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}...+\frac{1}{101}\right)\)

B = \(2\left(1-\frac{1}{101}\right)\)

B = \(2x\frac{100}{101}\)

B = \(\frac{200}{101}\)

\(B=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{100\cdot101}\)

\(B=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{100\cdot101}\right)\)

\(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(B=2\left(1-\frac{1}{101}\right)\)

\(B=2\cdot\frac{100}{101}\)

\(B=\frac{200}{101}\)

A=1x2+2x3+3x4+4x5+......+99x100+100x101

3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98)+100x101x(102-99)

3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100+100x101x102-99x100x101

3A=(1x2x3+2x3x4+3x4x5+4x5x6+...+99x100x101+100x101x102)-(0x1x2+1x2x3+2x3x4+3x4x5+...+98x99x100+99x100x101)

3A=100x101x102

A=100x101x102:3

A=343400

A = 1x2 + 2x3 + 3x4 + 4x5 + ... + 99x100 + 100x101

3A = 1x2x(3-0) + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) + 100x101x(102-99)

3A = 1x2x3 - 0x1x2 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100 + 100x101x102 - 99x100x101

3A = 100x101x102 - 0x1x2

3A = 100x101x102

A = 100x101x34

A = 343400

a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)

\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)

\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)^(a)

Đặt \(M=\left(1^2+2^2+........+100^2\right)\)

\(\Rightarrow M=1.1+2.2+.....+100.100\)

\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)

\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)

\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)

\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)⁽¹⁾

Đặt \(N=1.2+2.3+....+100.101\)

\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)

\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)

\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)

\(\Rightarrow3N=100.101.102-0\)

\(\Rightarrow N=343400\)

Thay N = 343400 vào 1) ta được:

M = 343400 - 5050 

=> M = 338350

Thay M = 338350 Vào (a) ta được:

A = 338350 . \(\frac{100}{101}\)

=> \(A=\frac{33835000}{101}\)

Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)

b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)

\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)

Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)

\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)

Rồi bạn làm như ở phần a) ý