1/2+1/12+1/30+...+1/9120+1/9506+1/9900. / 50-50/51-51/52-...-97/98-98/99-99/100
NT
Những câu hỏi liên quan
Tính
A=1+3^1-3^2+3^3-3^4+...........+3^99-3^100
B=50.(51^99+51^98+51^97+......+51^1+1)+1
Tính tổng
1/ 1 + (-2) + 3 + (-4) + . . . + 17 + (-18)
2/ -1 + 2 - 3 + 4 + . . . - 99 + 100
3/ 2 – 4 + 6 – 8 + . . . + 48 – 50 + 51 - 52
4/ – 1 + 3 – 5 + 7 - . . . . + 97 – 99 + 101
5/ 1 + 2 – 3 – 4 + . . . . + 97 + 98
1/ 1 + (-2) + 3 + (-4) + . . . + 17 + (-18)
= [1 + (-2)] + [3 + (-4)] + . . . + [17 + (-18)]
= (-1) + (-1) + . . . + (-1)
18 : 2 = 9
= (-1) . 9
= -9
2/ -1 + 2 - 3 + 4 + . . . - 99 + 100
= (-1 + 2) + (-3 + 4) + . . . (-99 + 100)
= 1 + 1 + . . . + 1
100 : 2 = 50
= 1 . 50
= 50
3/ 2 – 4 + 6 – 8 + . . . + 48 – 50 + 51 - 52
= (2 – 4) + (6 – 8) + . . . + (46 - 48) + (– 50 + 51 - 52)
= (-2) + (-2) + . . . + (-2) + (-51)
(48 - 2) : 2 + 1 = 24 : 2 = 12
= (-2) . 12 + (-51)
= (-24) + (-51)
= -75
4/ – 1 + 3 – 5 + 7 - . . . . + 97 – 99 + 101
= (–1 + 3) + (–5 + 7) + . . . . + (-93 + 95) + (97 – 99 + 101)
= 2 + 2 + . . . . + 2 + 99
(95 - 1) : 2 + 1 = 48 : 2 = 24
= 2 . 24 + 99
= 48 + 99
= 147
5/ 1 + 2 – 3 – 4 + . . . . + 97 + 98
= (1 + 2 – 3 – 4) + . . . . + (93 + 94 - 95 - 96) + (97 + 98)
= (-4) + . . . . + (-4) + 195
96 : 4 = 24
= (-4) . 24 + 195
= -96 + 195
= 99
1/ 1 + (-2) + 3 + (-4) + . . . + 17 + (-18)
=(1+3+5+..+17)+[-2+(-3)+(-4)+..+(-18)
goi ben tong so duong la A va tong so am la B ta co
A co ssh la :(17-1):2+1=9
B co ssh la :[-18-(-2)]:2+1=9
=>A=(17+1).9:2=81
=>B=[-18+(-2)].9:2=-72
=> A+B= 81+(-72)=9
vay tong bang 9
Tính: \(B=\frac{100^2+1^2}{100\cdot1}+\frac{99^2+2^2}{99\cdot2}+\frac{98^2+3^2}{98\cdot3}+...+\frac{52^2+49^2}{52\cdot49}+\frac{51^2+50^2}{51\cdot50}\)
1*99+2*98+3*97+...+51*50+50*50
tính A=1*99+2*98+3*97+...+49*51+50*50
Chứng Minh:
1/1*2+1/3*4+1/5*6+...+1/97*98+1/99*100=1/51+1/52+1/53+...+1/99+1/100
Tính:
a)1*4*7+4*7*10+7*10*13+....+100*103*106
b)1*4+4*7+7*10+.....+100*103
c)1*1*1+4*4*4+7*7*7+....+99*99*99
d)1*3*3*3+3*5*5*5+5*7*7*7+.....+49*51*51*51
e)1*99+2*98+3*97+......+50*50
f)1*99+3*97+5*95+....+49*51
Giúp mình nhé!
((1002+12)/(100*1))+((992+22)/(99*2))+((982+32)/(98*3))+...+((512+502)/(51*50))
cho A=1/11+1/12+1/13+1/14+...+1/50
so sánh A với 1/2
cho B=1/50+1/51+1/52+...+1/98+1/99
chứng minh rằng b <1/2
cho C=1/10+1/11+1/12+...+1/99+1/100
chứng tỏ C >1
a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
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