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Số số hạng của 12 + 22 + 32 + .. + 202 : (202 - 1) : 10 + 1 = 20
Số số hạng của 1 + 2 + 3 + .. + 20 : (20 - 1) + 1 = 20
S = {(12+ 202) - (1 + 20)} x 20 : 2 = 1930
Lúc nãy nhầm làm tổng của 2 dãy
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2:
a: =4+3/8+5+2/3
=9+3/8+2/3
=216/24+9/24+16/24
=216/24+25/24
=241/24
b; =2+3/8+1+1/4+3+6/7
=6+3/8+1/4+6/7
=6+5/8+6/7
=419/56
c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)
=6+3/8-1/4+1/3
=6+1/8+1/3
=6+11/24
=155/24
d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)
=3+13+5/6
=16+5/6
=101/6
e: =3+1/2+4+5/7-5-5/14
=3+4-5+1/2+5/7-5/14
=2+7/14+10/14-5/14
=2+12/14
=2+6/7=20/7
f: =9/2+1/2:11/2
=9/2+1/11
=99/22+2/22=101/22
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a: \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-14=6\)
b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)
\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)
\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)
c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)
\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)
\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)
d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)
\(=411-\left[\dfrac{110}{5}-4\right]\)
=410-22+4
=410-18
=392
e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)
\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)
\(=450-5\cdot\left[9\cdot8-12\right]+18\)
=468-5*60
=468-300
=168
f:
\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)
\(=102-150:\left[18-2\cdot4\right]+1\)
\(=103-\dfrac{150}{18-8}=103-15=88\)
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\(F=2^2+4^2+...+20^2\)
\(=\left(1.2\right)^2+\left(2.2\right)^2+...+\left(2.10\right)^2\)
\(=1.2^2+2^2.2^2+...2^2.10^2\)
\(=2^2\left(1+2^2+...+10^2\right)\)
\(=2^2.385\)
\(=4.385\)
\(=1540\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=C*[13*37*(5*3-15)]=0
\(A=\dfrac{2^{10}\cdot78}{2^8\cdot26\cdot4}=\dfrac{78}{26}=3\)
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A = - ( 1+2+3 +....+ 202) = - 203. 101 = -20503
B= ( 1+2-3-4) + ( 5+6-7-8) +..........+( 97+98 -99-100) + ( 101+102)
= -4 + (-4) .........+ (-4) + 203
= -4 .25 + 203 = 103
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Lời giải:
\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)
\(=4A=4.385=1540\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( 1 + 2 + .... + 99 + 100 ) . ( 12 + 22 + .... + 102) . ( 65 . 111 - 13 . 15 . 37)
=( 1 + 2 + .... + 99 + 100 ) . ( 12 + 22 + .... + 102) . ( 13. 5 . 3. 37 - 13 . 15
37)
=( 1 + 2 + .... + 99 + 100 ) . ( 12 + 22 + .... + 102) . ( 13. 15 . 37 - 13 . 15
37)
=( 1 + 2 + .... + 99 + 100 ) . ( 12 + 22 + .... + 102) . 0
=0
\(\left(1+2+...+100\right)\cdot\left(1^2+2^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot15\cdot37\right)\)
\(=\left[65\cdot111\left(1-1\right)\right]\cdot\left(1+2+...+100\right)\cdot\left(1^2+2^2+...+10^2\right)\)
=0
= \(\frac{20.21:2+2870}{2}=\frac{210+2870}{2}=\frac{3080}{2}=1540\)
\(=\frac{1\left(1+1\right)}{2}+\frac{2\left(2+1\right)}{2}+\frac{3\left(3+1\right)}{2}+...+\frac{20\left(20+1\right)}{2}\)
\(=\frac{1+1+2.2+2+3.3+3+...+20.20+20}{2}\)
\(=\frac{\left(1+...+20\right)+\left(1.1+2.2+3.3+...+20.20\right)}{2}\)
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