Đặt A=1.2.3+2.3.4+...+111.112.113

Ta có:4A=(1.2.3+2.3.4+...+111.112.113).4

          4A=1.2.3.(4-0)+2.3.4.(5-1)+...+111.112.113.(114-110)

          4A=1.2.3.4+2.3.4.5-1.2.3.4+...+111.112.113.114-110.111.112.113

          4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+110.111.112.113-110.111.112.113+111.112.113.114

          4A=111.112.113.114

      =)  A= 111.112.113.114/4( chia 4)

C = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 111 - 112 + 113 + 114 - 115 ( có 115 số; 115 chia 4 dư 2)

C = 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (110 - 111 - 112 + 113) + 114 - 115

C = 1 + 0 + 0 + ... + 0 + 114 - 115

C = 1 + 114 - 115

C = 115 - 115

C = 0

nửa cộng nửa trừ à

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C = 1 + 2 - 3 - 4 + 5 + ... - 111 - 112 + 113 + 114 + 115 ( có 115 số; 115 : 4 dư 3)

C = 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (110 - 111 - 112 + 113) + 114 + 115

C = 1 + 0 + 0 + ... + 0 + 114 + 115

C = 1 + 114 + 115

C = 115 × 2

C = 230

áp dụng công thức \(\frac{n\left(n-1\right)}{2}\)

<=>\(\frac{114\cdot\left(114-1\right)}{2}\)

<=> A =6441

A=1+2-3-4+5+6-7-8+...-111-112+113+114

A=1+(2-3-4+5)+(6-7-8+9)+...+(110-111-112+113)+114

A=1+ 0           +0 +.........+0+114

A=115

 1+2-3-4+5+6-7-8+...+111-112+113+114+115

=1+(2-3-4+5)+(6-7-8+9)+....................................+(110-111-112+113)+114+115

=230

thanks