Ta có : 

S=1.2+2.3+3.4+.............+n(n+1)  

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

=(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)  

Ta có các công thức:  

1^2 + 2^2 + 3^2 +...+ n^2

= n(n+1)(2n+1)/6  1 + 2 + 3 + ...+ n

= n(n+1)/2  

Thay vào ta có:  

S = n(n+1)(2n+1)/6 + n(n+1)/2  

=n(n+1)/2[(2n+1)/3 + 1]  

=n(n+1)(n+2)/3

****

3A = 1.2.3 + 2.3.3+......+n(n+1).3

= 1.2.3+2.3.(4-1)+.....+n(n+1)(n+2-n-1)

= 1.2.3 + 2.3.4-1.2.3+....+n(n+1)(n+2) - n(n+1)(n-1)

= n(n+1)(n+2)

=> A=  n(n+1)(n+2) / 3 

3A=1.2.3+2.3.3+3.4.3+...+n.(n+1).3 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1).n.[(n+1)-(n-2)]+n.(n+1).[(n+2)-(n-1)] 3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...-(n-2).(n-1).n+(n-1).n.(n+1)- (n-1).n.(n+1) + n.(n+1).(n+2) 3A=n.(n+1).(n+2) A=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

b) B = 2² + 4² + 6² + ... + 98² 

 \(\frac{1}{4}B=1^2+2^2+3^2+...+49^2\) 

\(\frac{1}{4}B=1+2\left(1+1\right)+3\left(2+1\right)+...+49\left(48+1\right)\) 

\(\frac{1}{4}B=1+2+1.2+2.3+3+...+48.49+49\) 

\(\frac{1}{4}B=\left(1+2+3+...+49\right)+\left(1.2+2.3+...+48.49\right)\) 

đặt A = 1.2 + 2.3 +...+ 48.49 ta có:

A = 1.2 + 2.3 +...+ 48.49

3A = 1.2.3 + 2.3.( 4 - 1) + ... + 48.49.( 50 - 47 )

3A = 1.2.3 + 2.3.4 - 1.2.3 +...+ 48.49.50 - 47.48.49

3A = 48.49.50

A = \(\frac{48.49.50}{3}=39200\)  

thay A = 39200 vào \(\frac{1}{4}B\) ta có:

\(\frac{1}{4}B=\left(1+2+3+...+49\right)+39200\) 

\(\frac{1}{4}B=1225+39200\)

 \(\frac{1}{4}B=40425\) 

B = 40425.4

B = 161700

vậy B = 161700

3A=1.2.3+2.3.4+3.4.3+.......+99.100.3

3A=1.2.(3-0) + 2.3 (4-1) + 3.4 . (5-2)+.......+ 99.100(101-98)

3A=(1.2.3+2.3.4+3.4.5+......+98.99.100)-(0.1.2+1.2.3+.....+98.99.100)

3A=99.100.101-0

3A=999900

A=999900:3

A=333300

em chưa học ẹ

A= 1.2+2.3+3.4.....+99.100

=>3A=1.2.3+2.3.3+3.3.4+....+99.100.3

=1.2(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=99.100.101-0

=999900

=>A=999900:3=333300

A = 1.2 + 2.3 + 3.4 + .. + 99.100

<=> 3A = 1.2.3 + 2.3.3 + 3.4.3 +...+ 99.100.3

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

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

            = 999900

<=> A = 999900 : 3 = 333300

A=1.2+2.3+3.4+...+99.100

3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3A=1.2.3+2.3.4+3.4.5+...+98.99.100+99.100.101 - 0.1.2-1.2.3-2.3.4-3.4.5-...-98.99.100

3A=99.100.101-0.1.2

3A=999900-0

3A=999900

A=999900:3

A=333300

Ta có : A = 1.2 + 2.3 + 3.4 + ...... + 100.101

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

=> 3A = 100.101.102

=> A = 100.101.102/3

=> A = 343400

sai rồi

A = 1.2 + 2.3 + 3.4 + ...... + 100.101

3A = 1.2.3 + 2.3.3 + 3.4.3 + ...... + 100.101.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ..... + 100.101.(102 - 99)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ...... + 100.101.102 - 99.100.101

3A = 100.101.102

A = 100.101.34

A = 343400

số số hạng : (99,100 -1.2) : 1.1 +1=90 số 

Tổng: (99.100 +1.2) x 90 : 2= 4513 ,5

\(A=1.2+2.3+3.4+....+99.100\\ 3.A=1.2.3+2.3.3+....+99.100.3\)

\(3.A=1.2.3+2.3.\left(4-1\right)+....+99.100\left(101-98\right)\\ 3.A=1.2.3+.....+99.100.101-98.99.100\)

\(3.A=99.100.101\\ A=33.100.101=333300\)