Đặt \(A=1.4+2.5+3.6+...+100.103\)

\(=1\left(2.2\right)+2\left(3+2\right)+3\left(4+2\right)+...+100\left(101+2\right)\)

\(=1.2+2.3+3.4+...+100.101+\left(1.2+2.2+3.2+...+100.2\right)\)

\(=1.2+2.3+3.4+...+100.101+2\left(1+2+3+...+100\right)\)

\(=1.2+2.3+3.4+...+100.101+2.100\left(100+1\right):2\)

\(=1.2+2.3+3.4+...+100.101+10100\)

Đặt \(B=1.2+2.3+3.4+...+100.101\)

\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+100.101.3\)

\(\Rightarrow3B=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)

\(\Rightarrow3B=100.101.102\)

\(\Rightarrow B=343400\)

Khi đó \(A=343400=10100=333300\)

Đặt A = 1.4 + 2.5 + 3.6 + 4.7 + ... + 100.103

3A = 3.(1.2 + 2.3 + 3.4 + ... + 100.101] + 3.(2 + 4 + 6 + ... + 200)

     = 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3 + 3.(2 + 4 + 6 + ... + 200)

\(\Rightarrow\) A  =  100.101.105:3 = 353500

Đặt A = 1.4 + 2.5 + 3.6 + ... + 100.103

= 1.(2 + 2) + 2.(3 + 2) + 3.(4 + 2) +.... + 100.(101 + 2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + (1.2 + 2.2 + 3.2 + ... + 100.2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 2(1 + 2 + 3 + .... + 100)

= 1.2 + 2.3 + 3.4 + .... + 100.101 + 2.100.(100 + 1) : 2

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 10100

Đặt B = 1.2 + 2.3 + 3.4 + .... + 100.101

=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + 100.101.3

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

=> 3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101

=> 3B = 100.101.102

=> B = 343400

Khi đó A = 343400 - 10100 = 333300

bạn tính kiểu khác đc ko ? kiểu ab mình ko hiểu lắm

nó giải vậy mà ko hiểu chịu lun

Ta thấy:

1.4 = 1.(1 + 3) = 1.(1 + 1 + 2) = 1.(1 + 1)+ 2.1

2.5 = 2.(2 + 3) = 2.(2 + 1 + 2) = 2.(2 + 1)+ 2.2

3.6 = 3.(3 + 3) = 3.(3 + 1 + 2) = 3.(3 + 1)+ 2.3

4.7 = 4.(4 + 3) = 4.(4 + 1 + 2) = 4.(4 + 1)+ 2.4

. . . . . . . . . . .

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

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + . . . + n(n + 1) + 2n

= 1.2 + 2 +2.3 + 4 + 3.4 + 6 + . . . + n(n + 1) + 2n

= [1.2 +2.3 +3.4 + . . . + n(n + 1)] + (2 + 4 + 6 + . . . + 2n)

Mà 1.2 + 2.3 + 3.4 + … + n.(n + 1) =\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Và 2 + 4 + 6 + . . . + 2n =\(\frac{\left(2n+2\right).n}{2}\)

=> C=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}+\frac{\left(2n+2\right).n}{2}-\frac{n.\left(n+1\right).\left(n+5\right)}{3}\)

hok tốt

Ta có : 

\(C=1.4+2.5+3.6+...+n\left(n+3\right)\)

\(\Rightarrow C=1\left(2+2\right)+2\left(3+2\right)+3\left(4+2\right)+...+n\left(n+1+2\right)\)

\(\Rightarrow C=1.2+1.2+2.3+2.2+3.4+3.2+...+n\left(n+1\right)+n.2\)

 \(\Rightarrow C=\left(1.2+2.3+3.4+...+n\left(n+1\right)\right)+2\left(1+2+3+...+n\right)\)

 \(\Rightarrow C=\frac{n\left(n+1\right)\left(n+2\right)}{3}+2\left(\frac{\left(n+1\right).n}{2}\right)\)  

\(\Rightarrow C=\frac{n\left(n+1\right)\left(n+2\right)}{3}+\left(n+1\right)n\)

~ 

C = 1.4+2.5+3.6+4.7+...+n.(n+3)

C= 1.(2+2) + 2.(3+2) + 3.(4+2) + 4.(5+2) + ...+n.[(n+1) + 2]

C = 1.2 + 1.2 + 2.3 + 2.2 + 3.4 + 3.2 + 4.5 + 4.2 + ...+ n.(n+1) + n.2

C = [(1.2+2.3+3.4+4.5+...+n.(n+1)] + ( 1.2+2.2+3.2+4.2+...+n.2)

Đặt A = 1.2 + 2.3 + 3.4+4.5 + ...+n.(n+1)

=>3A = 1.2.3+2.3.3+3.4.3+4.5.3+...+n.(n+1).3

3A = 1.2.(3-0)+2.3.(4-1) + 3.4.(5-2) + ...+n.(n+1).[(n+2) - (n-1)]

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + n.(n+1).(n+2) - (n-1).n.(n+1)

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

\(A=\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Đặt B = 1.2+2.2+3.2+4.2 +...+n.2

B = 2.(1+2+3+4+...+n)

B = 2. [(1+n).n:2]

B = 2. (1+n).n . 1/2

B = (1+n).n

Thay A;B vào C

có: \(C=\frac{n.\left(n+1\right).\left(n+2\right)}{3}+\left(1+n\right).n\)