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3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101
= 99.100.101
A=333300
B= (1.2 + 2.3 + 3.4 + ... + 100.101) - (1 + 2 + 3+ 4 + ... + 100)
= 333300 + 10100 - 5050
= 333300 + 5050
= 338350
A = 1*2 + 2*3 + 3*4 + ........+ 99*100
=>3A=1.2.3+2.3.3+3.4.3+...+99.100.3
<=> 3A =1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
<=> 3A =1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
<=> 3A = 99.100.101 = 999900
=> S = 333300
*) A=1+6+11+16+21+....+101
Dãy trên có: \(\left(101-1\right):5+1=21\)(số số hạng)
\(\Rightarrow A=\frac{\left(101+1\right)\cdot21}{2}=1071\)
*) Đặt C=\(1^2+2^2+3^2+....+98^2=1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\)
\(\Rightarrow B-C=\left(1\cdot2+2\cdot3+3\cdot4+....+98\cdot99\right)-\left(1\cdot1+2\cdot2+3\cdot3+....+98\cdot98\right)\)
\(=\left(1\cdot2-1\cdot1\right)+\left(2\cdot3-2\cdot2\right)+\left(3\cdot4-3\cdot3\right)+.....+\left(98\cdot99-98\cdot98\right)\)
\(=1\left(2-1\right)+2\left(3-2\right)+3\left(4-3\right)+....+98\left(99-1\right)\)
\(=1\cdot1+2\cdot1+3\cdot1+....+98\cdot1\)
\(=1+2+3+....+98\)
\(=\frac{\left(98+1\right)\cdot98}{2}=4851\)
A = 1 + 6 + 11 + 16 +21 +... + 101
Số chữ số của tổng A là :
( 101 - 1 ) : 5 + 1 = 21 (số)
Tổng A = 1 + 6 + ... + 101 = (101 + 1) . 21 : 2 = 1071
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)
\(=1.\frac{1}{101}=\frac{1}{101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)
\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)
\(=\frac{1}{101}\)
Ta có : \(\frac{x-1}{2}=\frac{x+1}{3}\)
<=> \(3\left(x-1\right)=2\left(x+1\right)\)
<=> \(3x-3=2x+2\)
<=> \(3x-2x=2+3\)
<=> x = 5
a, \(\frac{x-1}{2}=\frac{x+1}{3}\)
=> (x-1)3 = 2(x+1)
=> 3x - 3 = 2x + 2
=> 3x - 2x = 2 + 3
=> x = 5
b, \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}< 1\) (ĐPCM)
a) \(A=1.2+2.3+3.4+...+29.30\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+29.30\left(31-28\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+29.30.31-28.29.30\)
\(\Rightarrow3A=29.30.31\)
\(\Rightarrow A=29.30.31:3\)
\(\Rightarrow A=29.10.31\)
\(\Rightarrow A=8990\)
3A= 1.2.3+2.3.4+3.4.3 +......+ 29.30.3
3A= 1.2. ﴾3 ‐ 0﴿ + 2.3.﴾4 ‐ 1﴿ +3.4. ﴾5 ‐ 2﴿....... . 29.30. ﴾31 ‐ 28﴿
3A = ﴾1.2.3 + 2.3.4 + 3.4.5 +...... +18.20.21﴿ ‐ ﴾0.1.2 + 1.2.3 + 2.3.4 +.......+ 18.19.20﴿
3A = 29.30.31 ‐ 0.1.2
3A =26970‐0
3A= 26970
A=26970:3
A = 8990.
Vậy A=8990
\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)
\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)
\(=\frac{1}{1.101}\)
\(=\frac{1}{101}\)
k cho mk nha
a) A=(100-1):1+1=100 số hạng
A=100:2=50 cặp
tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101
tính giá trị của biểu thức A: 50*101=5050
[ mình tính theo công thức đó ]
Ta có: $a=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots+\dfrac{1}{99\cdot100}\right)+1+\dfrac{1}{100}$
Nhận xét: $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$
Do đó: $\dfrac{1}{1\cdot2}=\dfrac{1}{1}-\dfrac{1}{2}$
$\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}$
$\cdots$
$\dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}$
=> $\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\cdots+\dfrac{1}{99\cdot100} =1-\dfrac{1}{100}$
Do đó:
$a=1-\dfrac{1}{100}+1+\dfrac{1}{100}$ $=2$
Ta tính: $\dfrac{(1+a^2)^3+3}{2}+3$
$=\dfrac{(1+2^2)^3+3}{2}+3$
$=\dfrac{(1+4)^3+3}{2}+3$
$=\dfrac{5^3+3}{2}+3$
$=\dfrac{125+3}{2}+3$
$=\dfrac{128}{2}+3$
$=64+3$
$=67$
[(1+2/2)×3+3]:2+3=(2×3+3):2+3=9:2+3=9/2+3=15/2