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3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3
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D = 1.2 + 2.3+ 3.4 +...+ 99.100
=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=999900
=>D=999900:3=333300
Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)
=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]
=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)
=n.(n+1).(n+2)-0.1.2
=n.(n+1)(n+2)
=>Dn=n.(n+1)(n+2):3
=>điều cần chứng minh
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Em tham khảo:
Câu hỏi của nguyễn huy bảo - Toán lớp 7 - Học trực tuyến OLM
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Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101
=> 3S = 99.100.101
=> S = \(\frac{99.100.101}{3}=333300\)
ta xét
\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)
\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)
\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)
\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)
Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi a1=1.2=>3a1=1.2.3=>3a1=1.2.3-0.1.2
a2=2.3=>3a2=2.3.3=>3a1=2.3.4-1.2.3
tương tự .....
đến an-1=(n-1)n=>3an-1=3(n-1)n =>an-1=(n-1)n(n+1)-(n-2)(n-1)n
an=n.(n+1)=>3an=3.n(n+1)=>3an=n(n+1)(n+2)-(n-1)n(n+1)
cộng từng vế các đẳng thức ta được:
3a1+3a2+....+3an-1+3an=1.2.3-0.1.2+2.3.4-1.2.3+....+n(n+1)(n+2)-(n-1)n(n+1)
3(a1+a2+...+an-1+an)=n(n+1)(n+2)
3(1.2+2.3+3.4+...+n(n+1)=n(n+1)(n+2)
=>1.2+2.3+3.4+...+n.(n+1)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
đặt\(A=1.2+2.3+3.4+.......+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+....+n\left(n+1\right).3\)\(=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+n\left(n-1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+....+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
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3s=1.2(3-0)+2.3(4-1)+...+n.(n+1).(n+2)-(n-1)
3a=1.2.3+2.3.4+3.4.5+....+(n-1).n(n+1)+n(n+1)(n+2)-
0.1..2-1.2.3-3.4.5-....-(n-1)n(n+1)
3a=n.(n+1)(n+2)
a=n(n+1)(n+2)/3
1.2+2.3+3.4.....+n.(n+1)=A
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
ai mình nt mik lại
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3S2=1*2*(3-0)+2*3*(4-1)+...+ n*(n+1)*[(n+2)-(n-1)]
3S2=1*2*3+2*3*4+...+n*(n+1)*(n+2)-0*1*2-1*2*3-...-(n-1)*n*(n+1)
3S2=n*(n+1)*(n+2)
\(S_2=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)