a) 1 + (-2) + 3 + (-4) + .. + 19 + (-20) = (-1) + (-1) + ... + (-1) (có 10 số -1) = (-1) . 10 = -10 b) 1 - 2 + 3 - 4 + ... + 99 - 100 = (-1) + (-1) + ... + (-1) (có 50 số -1) = (-1) . 50 = -50 c) 2 - 4 + 6 - 8 + ... + 48 - 50 = (-2) + (-2) + ... + (-2) (có 25 số -2) = (-2) . 25 = -50 d) -1 + 3 - 5 + 7 - ... + 97 - 99 = (-1) + (-2) + (-2) + ... (-2) (có 49 số -2) = (-1) + (-2) . 49 = (-1) + (-98) = -99 e) 1 + 2 - 3 - 4 + ... + 97 + 98 - 99 - 100 = 1 + 2 - 3 - 4 + ... + 97 + 98 - 99 - 100 + 101 (ta cộng thêm 101 cho dễ tính) = 1 + (2 - 3 - 4 + 5) + ... + (98 - 99 - 100 + 101) = 1 + 0 + ... + 0 = 1 - 101 (ta bớt 101 để ra kết quả vì lúc nãy thêm 101) = -100 \(A=1\left(2+2\right)+2\left(2+3\right)+3\left(2+4\right)+.....+\left(n-1\right)\left(2+n\right)\) \(\Leftrightarrow A=1.2+1.2+2.3+2.2+3.4+2.3+....+\left(n-1\right)n+2\left(n-1\right)\) \(\Leftrightarrow A=\left(1.2+2.3+.....+\left(n-1\right)n\right)+2\left(1+2+3+....+\left(n-1\right)\right)\) Giả sử A=B+C Với \(\begin{cases}B=1.2+2.3+.....+\left(n-1\right)n\\C=2\left[1+2+....+\left(n-1\right)\right]\end{cases}\) Ta có \(3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]\) \(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3+.....+\left(n-1\right)n\left(n+1\right)-\left(n-2\right)\left(n-1\right)n\) \(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)}{3}\) Mặt khác \(C=2\left[1+2+....+\left(n-1\right)\right]\) \(\Rightarrow C=2.\frac{\left[\left(n-1\right)+1\right]n}{2}=n^2\) \(\Rightarrow A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\) Vậy \(A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\) D =1.4+2.5+3.6+.......+99.102 D = 1. (2+2) +2.(2+3) +3.(2+4)+...+99.(100+2) D = 1.2+1.2+2.2+2.3+2.3+3.4+...+2.99+99.100 D = (1.2+2.3+3.4+...+99.100) +2.(1+2+3+4+...+99) *Gọi A= 1.2+2.3+3.4+...+99.100 3A = 3.(1.2+2.3+3.4+...+99.100) 3A = 1.2.3+2.3.3+...+99.100.3 3A = 1.2.3 +2 .3.(4-1)+...+99.100.(101-98) 3A = 1.2.3+2.3.4-1.2.3+...+ 99.100.101-98.99.100 3A = 99.100.101 3A = 3.33.100.101 A = 33.100.101 A = 333300 * Gọi B = 2. (1+2+3+4+...+99) \__có 99 số hạng ___/ B= 2.[(1+99).99:2] B = 2 .4950 B = 9900 A+B = 333300+9900 =343200 Vậy D =343200 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 |