a)(1+2+3+....+100)x(12+22+32+....+102)x(65x111-13x15x37)
b)276:910
c)410x8
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H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
CM
26 tháng 1 2018
Ta có 12 + 22 + 32 + …102 = 385
Suy ra ( 12 +22 + 32 +…+102 ) .32 = 385.32
Do đó ta tính được A = 32 + 62 + 92 + …+302 = 3465
AH
Akai Haruma
Giáo viên
2 tháng 1 2021
Lời giải:
\(B=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2=2^2(1^2+2^2+...+10^2)\)
\(=4A=4.385=1540\)
PA
18 tháng 7 2016
a.
\(\frac{2^{10}\times13+2^{10}\times65}{2^8\times104}=\frac{2^{10}\times\left(13+65\right)}{2^8\times104}=\frac{2^2\times78}{104}=\frac{4\times78}{104}=\frac{312}{104}=3\)
b.
\(\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(65\times111-15\times37\times13\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times\left(7215-7215\right)\)
\(=\left(1+2+...+100\right)\times\left(1^2+2^2+...+10^2\right)\times0\)
= 0
HS
1
BD
19 tháng 6 2023
\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)
Gọi 1+2+3+...+10 là P
Số số hạng là: (10 - 1) : 1 +1 = 10 (số)
P = (10+1) . 10 : 2 = 55
P = 55
Gọi \(1\cdot2+2\cdot3+....+9\cdot10\) là C
\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)
\(=>A=P+C\\ =>A=55+330\\ A=385\)
b)
\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)
\(\left(1+2^2+3^2+....+10^2\right)=A\)
\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)
MV
10 tháng 6 2017
\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\\ =\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\\ =6+6+6+...+6\left(10\text{ số }6\right)\\ =6\cdot10\\ =60\)
\(b,3+\left(-12\right)+13+\left(-22\right)+23+\left(-32\right)+...+93+\left(-102\right)\\ =\left[3+\left(-12\right)\right]+\left[13+\left(-22\right)\right]+\left[23+\left(-32\right)\right]+...+\left[93+\left(-102\right)\right]\\ =\left(-9\right)+\left(-9\right)+\left(-9\right)+...+\left(-9\right)\left(10\text{ số }-9\right)\\ =\left(-9\right)\cdot10\\ =-90\)
10 tháng 6 2017
\(a,\left(-5\right)+11+\left(-15\right)+21+\left(-25\right)+31+...+\left(-95\right)+101\)
\(=\left[\left(-5\right)+11\right]+\left[\left(-15\right)+21\right]+\left[\left(-25\right)+31\right]+...+\left[\left(-95\right)+101\right]\)
\(=6+6+6+...+6\) (10 số 6)
\(=6.10=60\)
\(\)
