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-1+3-5+7-....+97-99
= (-1-99)+(-3-97)+. . .+(-4 9-51)
= (-100)+(-100)+. . .+(-100)
Có 50 cặp -100
Nên Tổng bằng : -100.50= -5000
Vậy. . . .= -5000
a) -1+3-5+7-...+97-99
=(-1-99)+(-3-97)+...+(-49-51) (có tất cả 25 cặp)
=(-100)+(-100)+...+(-100)
=(-100).25=-2500
KILL !!!! Team Free Fire ~_~
Bạn làm sai rồi nha! Có tất cả 50 số hạng. Nhóm mỗi nhóm 2 số hạng thì có 50:2=25 cặp chứ bạn
b) 1+2-3-4+...+97+98-99-100
=(1+2-3-4)+(5+6-7-8)+...+(97+98-99-100) (có tất cả 25 cặp)
=(-4)+(-4)+...+(-4)
=(-4).25=-100
Đặt C=1.3.5.7...99
Đặt D=51/2.52/2.53/2 ....100/2
Ta có:C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
Vậy C=D
Ta có :
\(1.3.5.....99=\frac{\left(1.3.5.....99\right)\left(2.4.6.....98\right)}{2.4.6.....98}=\frac{1.2.3.....99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.....100}{2.2.2.....2}\)
\(=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)
Vậy......................
~ Hok tốt ~
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\frac{4949}{9900}\)
\(A=\frac{4949}{19800}\)
bài 1 :
a) S1=( 1 + 3 - 5 - 7 )+(9+11-13-15)+...+(393+395-397-399)
S1=(-8)+(-8)+...+(-8)
S1=(-8)*199
S1=-1592
b)S2=(1-2-3+4)+( 5 - 6 - 7 +8)+...+( 97 - 98 - 99 + 100)
S2=0+0+...+0
S2=0*100
S2=0
phần c và d tương tự nhé
BÀI 2
c)<=>2(x-1)+4 chia hết x-3
=>8 chia hết x-3
=>x-3\(\in\){-1,-2,-4,-8,1,2,4,8}
=>x\(\in\){2,1,-1,-5,4,5,7,11}
5151=5150.51=(512)25.51=260125.51=..........01.51=.............51
Xét tử ta có:
\(101+100+99+98+...........+3+2+1\)
\(=1+2+3+..........+99+100+101\)
\(=\frac{101.102}{2}=5151\)
Xét mẫu ta có:
\(101-100+99-98+.......+3-2+1\)
\(=\left(101-100\right)+\left(99-98\right)+.......+\left(3-2\right)+1\)
\(=1+1+.......+1+1=51\)
\(\Rightarrow A=\frac{5151}{51}=101\)
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- Series 1: 1+23+299+21011 + \frac{2}{3} + \frac{2}{99} + \frac{2}{101} with 100
1+23+299+21011 + \frac{2}{3} + \frac{2}{99} + \frac{2}{101}Since the notation is a bit unclear, I'll assume you meant to add those specific terms together:
Let's compute each term:
1=1.0001 = 1.000 23≈0.667\frac{2}{3} \approx 0.667 299≈0.0202\frac{2}{99} \approx 0.0202 2101≈0.0198\frac{2}{101} \approx 0.0198Adding them together:
1+0.667+0.0202+0.0198≈1.7071 + 0.667 + 0.0202 + 0.0198 \approx 1.707- Series 2: 99100+99101+…+99199\frac{99}{100} + \frac{99}{101} + \ldots + \frac{99}{199} with 2
∑n=10019999n≈99∫1001991ndn\sum_{n=100}^{199} \frac{99}{n} \approx 99 \int_{100}^{199} \frac{1}{n} dn =99[ln(n)]100199= 99 [\ln(n)]_{100}^{199} =99(ln(199)−ln(100))= 99 (\ln(199) - \ln(100)) ≈99(ln(199)−ln(100))\approx 99 (\ln(199) - \ln(100)) =99(ln(199)−ln(100))≈99(5.293−4.605)≈99×0.688≈68.112= 99 (\ln(199) - \ln(100)) \approx 99 (5.293 - 4.605) \approx 99 \times 0.688 \approx 68.112This series seems to add the fraction 99n\frac{99}{n} for nn going from 100 to 199. We can calculate the approximate sum using an integral approximation:
- Series 3: 1451+1452+…+14100\frac{14}{51} + \frac{14}{52} + \ldots + \frac{14}{100} with 14
∑n=5110014n≈14∫511001ndn\sum_{n=51}^{100} \frac{14}{n} \approx 14 \int_{51}^{100} \frac{1}{n} dn =14[ln(n)]51100= 14 [\ln(n)]_{51}^{100} =14(ln(100)−ln(51))= 14 (\ln(100) - \ln(51)) ≈14(4.605−3.933)≈14×0.672≈9.408\approx 14 (4.605 - 3.933) \approx 14 \times 0.672 \approx 9.408This series adds the fraction 14n\frac{14}{n} for nn going from 51 to 100. Using integral approximation:
- Series 4: 01+12+23+…+9899+99100\frac{0}{1} + \frac{1}{2} + \frac{2}{3} + \ldots + \frac{98}{99} + \frac{99}{100} with 2
∑n=099nn+1\sum_{n=0}^{99} \frac{n}{n+1}Let's evaluate the sum from 0/1 to 99/100:
This can be simplified using properties of telescoping series.
- Series 5: 150+251+…+5099+51100\frac{1}{50} + \frac{2}{51} + \ldots + \frac{50}{99} + \frac{51}{100} with 4
∑n=50100nn+49\sum_{n=50}^{100} \frac{n}{n+49}Similarly, this series can be evaluated using:
Each term for the series should be computed individually or approximated using integrals for a close sum. Let me know if you need further breakdowns for specific series!
Can you explain the steps for evaluating these series?Sure, I'd be happy to walk you throug