tính A = 1x2 + 2x3 +3x4 +... +n x (n +1)
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a)
Số số hạng của dãy trên là;
(n - 1) : 1 + 1 = n(số hạng)
Tổng dãy trên là:
(n + 1) x n : 2 = ? (tùy giá trị n)
b) Đặt A = 1x2 + 2x3 + 3x4 + ... + 99 x 100
3A= 3 x ( 1x2 + 2x3 + 3x4 + ... + 99 x 100)
3A = 1 x 2 x (3 - 0) + 2 x 3 x(4-1) + .....+99.100.(101 - 98)
3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + .......+ 99.100.101
3A = 99.100.101
A = \(\frac{\text{99.100.101}}{3}=333300\)
a, 1 + 2 + 3 + ... + n
= ( 1 + n) × n : 2
b, 1×2 + 2×3 + 3×4 + ... + 99×100
= 1/3 × ( 1×2×3 + 2×3×3 + 3×4×3 + ... + 99×100×3)
= 1/3 × [ 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 99×100×(101-98) ]
= 1/3 × ( 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 99×100×101 - 98×99×100 )
= 1/3 × [ ( 1×2×3 + 2×3×4 + 3×4×5 + ... + 99×100×101) - ( 0×1×2 + 1×2×3 + 2×3×4 + ... + 98×99×100) ]
= 1/3 × ( 99×100×101 - 0×1×2)
= 1/3 × ( 99×100×101 - 0)
= 1/3 × 99×100×101
= 333 300
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1x2+2x3+3x4+...+nx(n+1)
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + n.(n + 1).[(n + 2).(n - 1)]
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n + 1).(n + 2)
=> 3A = n.(n + 1).(n + 2)
=> A = n.(n + 1).(n + 2) / 3
Cách làm mk làm giống Edokawa Conan nhé kw ;\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{n\left(n+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
\(=\frac{n+1}{n+1}-\frac{1}{n+1}\)
\(=\frac{n}{n+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
program tinhtoan;
uses crt;
var: i;n:interger;
S:real;
writeln(' Nhap n='); readln(n);
S:=0;
For i:=1 to n*(n*1) do S:=S+\(\frac{1}{i};\)
writeln(' S=',S);
End.
(ps: ko chắc )
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = 1 x 2 + 2 x 3 + 3 x 4 + ... + n x ( n - 1)
=> 3A = 1 x 2 x (3 - 0) + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + ... + n x (n - 1) x [(n + 2) x (n + 1)]
=> 3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + ... + n x (n + 1) x (n + 2)
=> 3A = n x (n + 1) x (n + 2)
=> A = n x (n + 1) x (n + 2) / 3
3S=1.2.3+3.4.5+...+n.(n-1).3
1.2.(3-0).......................................................
k mk đi mk giải tiếp cho nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1\times2+2\times3+3\times4+...+19\times20\)
\(A\times3=3\times\left(1\times2+2\times3+3\times4+...+19\times20\right)\)
\(A\times3=1\times2\times3+2\times3\times3+3\times4\times3+...+19\times20\times3\)
\(A\times3=1\times2\times3+2\times3\times\left(4-1\right)+3\times4\times\left(5-2\right)+....+19\times20\times\left(21-18\right)\)
\(A\times3=1\times2\times3-1\times2\times3+2\times3\times4-2\times3\times4+3\times4\times5+...+19\times20\times21\)
\(A\times3=\left(1\times2\times3-1\times2\times3\right)+\left(2\times3\times4-2\times3\times4\right)+...+\left(18\times19\times20-18\times19\times20\right)+19\times20\times21\)
\(A\times3=19\times20\times21\)
\(A\times3=7980\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2013}{2014}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{n+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{2014}\)
\(\Rightarrow n+1=2014\)
\(\Rightarrow n=2014-1\)
\(\Rightarrow n=2013\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
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\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)