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

Đề có thiếu ko vậy bạn

Tính nhanh:

2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3

\(S=1.4+2.5+3.6+4.7+...+n\left(n+3\right)\)

\(S=4+10+18+21+...+n\left(n+3\right)\)

S gồm có :

\(\dfrac{n\left(n+3\right)-4}{4}+1\) ( số hạng )

Tổng S là:

\(S=\left[n\left(n+3\right)+4\right].\left[\dfrac{n\left(n+3\right)-4}{4}+1\right]:2\)

\(S=\left(n^2+3n+4\right)\left[\dfrac{n^2+3n-4}{4}+1\right].\dfrac{1}{2}\)

\(S=\dfrac{n^2+3n+4}{2}.\dfrac{n^2+3n}{4}\)

1 số hạng nha mình đang tìm cách giải thích

em xin chịu

Ta có: A=1.4+2.5+3.6+.....+98.101+99.102

          A=1.(2+2)+2.(3+2)+3(4+2)+........+98.(100+1)+99(100+2)

          A=1.2+1.2+2.3+2.2+3.4+3.2+.........+98.100+98.1+99.100+99.2

          A=(1.2+2.3+3.4+.......+99.100)+2.(1+2+3+4+......+99)

          A=...................

TK

S=1.4+2.5+3.6+4.7+....+n.(n+3) S = 1. ( 2 + 2 ) + 2. ( 3 + 2 ) + 3. ( 4 + 2 ) + . . . + n . [ ( n + 1 ) + 2 ] S = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) + ( 1.2 + 2.2 + 3.2 + . . . . + n .2 ) Đặt A = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) 3 A = 1.2.3 + 2.3. ( 4 − 1 ) + . . . . + n . ( n + 1 ) . [ ( n + 2 ) − ( n − 1 ) 3 A = 1.2.3 + 2.3.4 − 1.2.3 + . . . . + n . ( n + 1 ) . ( n + 2 ) − ( n − 1 ) . n . ( n + 1 ) 3 A = n . ( n + 1 ) . ( n + 2 ) A = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. ( 1 + 2 + 3 + . . . + n ) S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. n . ( n + 1 ) : 2 S = n . ( n + 1 ) . ( n + 2 ) : 3 + n . ( n + 1 ) S = n . ( n + 1 ) . [ ( n + 2 ) : 3 + 1 )

D = 1^2 + 2^2 + 3^2 + ... + n^2 
   = 1.( 2 - 1 ) + 2.( 3-1 ) + 3.( 4-1 ) + .... + n.[ ( n+ 1) - 1 ]
   = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + .... + n.( n+1 ) - n

   = [ 1.2 + 2.3 + 3.4 + ..... + n.( n + 1 ) ] - ( 1 + 2 + 3 + .... + n ) 
   = { [ n.( n+1 ).( n+2 )] /3 } - { [ n.( n+1)] /2 } 
   = { n(n+1)(2n+1) }/ 6 
Vậy......... 

TK

TK

D=12+22+32+...+n2