a) Ta có : \(C=1\cdot2+2\cdot3+...+100\cdot101\)
\(\Rightarrow3C=1\cdot2\cdot3+2\cdot3\cdot3+...+100\cdot101\cdot3\)
\(\Rightarrow3C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+100\cdot101\left(102-99\right)\)
\(\Rightarrow3C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+100\cdot101\cdot102-99\cdot100\cdot101\)
\(\Rightarrow3C=100\cdot101\cdot102\)
\(\Rightarrow C=\frac{100\cdot101\cdot102}{3}\) = 343400
b) Ta có : \(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot3+...+n\left(n+1\right)\cdot3\)
\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n-1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+2\cdot3\cdot4-3\cdot4\cdot5+...+n\left(n+1\right)\left(n+2\right)-n\left(n+1\right)\left(n-1\right)\)
\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
nhân biểu thức này với 3 là được mà!!!!!!!!!!!!!!!!!!!!!!!!!
a ) Ta có 3C = 1.2.3 + 2.3.3 + .... + 100.101.3
= 1.2.3 + 2.3.(4 - 1) + .... + 100.101(102 - 99)
=1.2.3 + 2.3.4 - 1.2.3 + .... + 100.101.102 - 99.100.101
= (1.2.3 - 1.2.3) + (2.3.4 - 2.3.4) + ... + (99.100.101 - 99.100.101) + 100.101.102
= 100.101.102
\(\Rightarrow C=\frac{100.101.102}{3}=100.101.34=343400\)
b ) ta có dạng tổng quát : A = 1.2 + 2.3 + ... + n(n + 1) = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Giải;
a) Ta có ;C=1.2+2.3+...+100.101
3C=1.2.3+2.3.3+.......+100.101.3
3C=1.2.3+2.3.(4-1)+...+100.101.(102-99)
3C=1.2.3+2.3.4-2.3.1+........+100.101.102
3C=100.101.102
Suy ra: C=100.101.102/3=343400
b) A=1.2+2.3+...+n(n+1)
Ta có:
3.A=1.2.(3-0)+2.3.(4-1)+3.4.(5 -2)...+ n.(n+1) . ((n+2) - (n-1))
3.A=1.2.3+2.3.4+3.4.5+...+ (n-1) . n. (n+1)+ n. (n+1). (n+2) -
0.1.2 -1.2.3 -2.3.4 -3.4.5 -...(n-1)n(n+1)
3A=n.(n+1).(n+2)
A=n.(n+1).(n+2)\3
