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Bạn dùng công thức này mà làm nhé :\(\frac{n}{a\times\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)
Ví dụ :\(\frac{2}{3\times5}=\frac{1}{3}-\frac{1}{5};\frac{3}{4\times7}=\frac{1}{4}-\frac{1}{7};\frac{5}{6\times11}=\frac{1}{6}-\frac{1}{11}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Mình chỉnh lại đề B nha:
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(M=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{46.51}\)
\(M=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+...+\frac{51-46}{46.51}\)
\(M=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{46}-\frac{1}{51}\)
\(M=1-\frac{1}{51}=\frac{50}{51}\)
\(N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{199\cdot201}\)
\(N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{199}-\frac{1}{201}\right)\)
\(N=\frac{1}{2}\cdot\left(1-\frac{1}{201}\right)\)
\(N=\frac{1}{2}\cdot\frac{200}{201}=\frac{100}{201}\)
\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
a)
Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:
2/9+6/27+8/36+12/54+16/72+18/81=
2/9+2/9+2/9+2/9+2/9+2/9=
2/9*6=
12/9=
4/3
Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3
b)
Ta có:
1-2/5=3/5
1-2/7=5/7
1-2/9=7/9
...
1-2/99=97/99
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=
3/5*5/7*7/9*...*97/99=
(3*5*7*...*97)/(5*7*9*...*99)=
3/99=
1/33
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33
c)
Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:
S=1/2+1/4+1/8+1/16+...+1/1024
S*2=1+1/2+1/4+1/8+...+1/512
S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)
S=1-1/1024
S=1023/1024
Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024
1) A=1-2+3-4+5-6+.....+99-100+101?
Giải
A=1-2+3-4+5-6+.....+99-100+101.
Ta viết lại tổng như sau:
A = 101 - 100 + 99 - 98 + ... + 5 - 4 + 3 - 2 + 1
A = 1 + 1 + ... + 1 + 1 + 1
Số phép trừ trong dãy tính là:
( 101 - 1 ) : 2 = 50 ( phép trừ )
Kết quả dãy số là:
1 x 50 + 1 = 51
Vậy:
A=1-2+3-4+5-6+.....+99-100+101.
A= 51
2) B=1+11+21+...+991
=(1+991)+(2+998)+...
=992 x 50
=4960
A=55*[(1/11*16) * 1/5 + (1/16*21) * 1/5 + (1/21*26) * 1/5 + (1/26*31) * 1/5 + (1/31*36) * 1/5 + (1/36*41) * 1/5 ]
= 55* [ (1/11*16) + (1/16*21) + (1/21*26) + (1/26*31) + (1/31*36) + (1/36*41) ] * 1/5
= 55 * [ 1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31+1/31-1/36+1/36-1/41) * 1/5
= 55 * ( 1/11 - 1/41 ) * 1/5
= 30/41
\(A=\frac{1}{1\times6}+\frac{1}{6\times11}+\frac{1}{11\times16}+...+\frac{1}{31\times36}\)
\(=\frac{1}{5}.\left(\frac{5}{1\times6}+\frac{5}{6\times11}+...+\frac{5}{31\times36}\right)=\frac{1}{5}\times\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(=\frac{1}{5}\times\left(1-\frac{1}{36}\right)=\frac{1}{5}\times\frac{35}{36}=\frac{7}{36}\)