A= 32 / 1.4 + 32/4.7+32/10.13+..........................+32/97.100
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28 tháng 11 2021
TK
S=1.4+2.5+3.6+4.7+....+n.(n+3) S = 1. ( 2 + 2 ) + 2. ( 3 + 2 ) + 3. ( 4 + 2 ) + . . . + n . [ ( n + 1 ) + 2 ] S = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) + ( 1.2 + 2.2 + 3.2 + . . . . + n .2 ) Đặt A = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) 3 A = 1.2.3 + 2.3. ( 4 − 1 ) + . . . . + n . ( n + 1 ) . [ ( n + 2 ) − ( n − 1 ) 3 A = 1.2.3 + 2.3.4 − 1.2.3 + . . . . + n . ( n + 1 ) . ( n + 2 ) − ( n − 1 ) . n . ( n + 1 ) 3 A = n . ( n + 1 ) . ( n + 2 ) A = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. ( 1 + 2 + 3 + . . . + n ) S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. n . ( n + 1 ) : 2 S = n . ( n + 1 ) . ( n + 2 ) : 3 + n . ( n + 1 ) S = n . ( n + 1 ) . [ ( n + 2 ) : 3 + 1 )
D = 1^2 + 2^2 + 3^2 + ... + n^2
= 1.( 2 - 1 ) + 2.( 3-1 ) + 3.( 4-1 ) + .... + n.[ ( n+ 1) - 1 ]
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + .... + n.( n+1 ) - n
= [ 1.2 + 2.3 + 3.4 + ..... + n.( n + 1 ) ] - ( 1 + 2 + 3 + .... + n )
= { [ n.( n+1 ).( n+2 )] /3 } - { [ n.( n+1)] /2 }
= { n(n+1)(2n+1) }/ 6
Vậy.........
DP
29 tháng 6 2017
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
ML
29 tháng 6 2017
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
28 tháng 6 2017
đây là toán lớp 5 cơ mà
a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)
A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)
A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))
A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)
A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))
A=\(\frac{1}{3}\)x\(\frac{18}{19}\)
A=\(\frac{6}{19}\)
DQ
1
9 tháng 5 2017
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)
9 tháng 5 2017
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)
\(A=\frac{3}{1}-\frac{3}{100}\)
\(A=\frac{297}{100}\)
27 tháng 4 2016
Ta thấy: 1/1-1/4 = 3/4 = 3.(1/1.4)
1/4-1/7 = 3/28 = 3.(1/4.7)
A = 3(1/1-1/4+1/4-1/7+...+1/97-1/100)
A = 3.(1-1/100)
A = 3.(99/100)
A = 297/100
27 tháng 4 2016
\(A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}\)
\(A=\frac{33}{100}\)
27 tháng 4 2017
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
2 tháng 5 2017
A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)
A:3=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\)
A:3=\(\frac{1}{1}-\frac{1}{100}\)
A:3=\(\frac{99}{100}\)
A=\(\frac{99}{100}.3\)
A=\(\frac{297}{100}\)
BB
2 tháng 5 2017
\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(A:3=\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A:3=\frac{1}{1}-\frac{1}{100}\)
\(A:3=\frac{99}{100}\)
\(A=\frac{99}{100}.3\)
\(A=\frac{297}{100}\)
HC
1
A=3.(\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\) )
A=3.(\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}\)\(+...+\frac{1}{97}-\frac{1}{100}\))
A=3.\(\left(1-\frac{1}{100}\right)\)=\(\frac{297}{100}\)
\(A=3\left(\frac{3}{1.4}\right)+3\left(\frac{3}{4.7}\right)+3\left(\frac{3}{7.10}\right)+...+3\left(\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}\right)+3\left(\frac{1}{4}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{10}\right)+...+3\left(\frac{1}{97}-\frac{1}{100}\right)\)
\(=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+...+\frac{3}{97}-\frac{3}{100}\)
\(=3-\frac{3}{100}\)
\(=\frac{297}{100}\)