tính tổng
a= 2/1x5+2/5x9+2/9x13+...+2/2001x2005
giúp em với ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`2/[1xx5]+2/[5xx9]+2/[9xx13]+....+2/[93xx97]+2/[97xx101]`
`=1/2xx(4/[1xx5]+4/[5xx9]+4/[9xx13]+....+4/[93xx97]+4/[97xx101])`
`=1/2xx(1-1/5+1/5-1/9+1/9-1/13+...+1/93-1/97+1/97-1/101)`
`=1/2xx(1-1/101)`
`=1/2xx100/101`
`=50/101`
Lời giải:
$12A=1.5.12+5.9.(13-1)+9.13(17-5)+13.17(21-9)+....+77.81(85-73)+81.85(89-77)$
$=60+(5.9.13+9.13.17+13.17.21+...+77.81.85+81.85.89)-(1.5.9+5.9.13+9.13.17+...+73.77.81+77.81.85)$
$=60+81.85.89 - 1.5.9=612780$
A = 1.5 + 5.9 + 9.13 + ... + 81.85
A = \(\dfrac{12}{12}\)(1.5 + 5.9 + 9.13 + 81.85)
A = \(\dfrac{1}{12}\).(1.5.12 + 5.9.12.+ 9.13.12 + ...+ 81.85.12]
A = \(\dfrac{1}{12}\).[1.5.(9 + 3) + 5.9.(13 - 1) + 9.13.(17 - 5) +...+ 81.85.(89 - 77)]
A = \(\dfrac{1}{12}\).[1.5.9 + 1.3.5 + 5.9.13 - 5.9.1 + 9.13.17 - 9.13.5 + ...+ 81.85.89 - 81.85.77]
A = \(\dfrac{1}{12}\).[1.3.5 + 81.85.89]
A = \(\dfrac{1}{12}\).[15 + 612765]
A = \(\dfrac{1}{12}\).612780
A = 51065
\(I=\dfrac{2}{1\times5}+\dfrac{2}{5\times9}+\dfrac{2}{9\times13}+...+\dfrac{2}{181\times185}\)
\(=\dfrac{1}{2}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{181\times185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{181}-\dfrac{1}{185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{185}\right)=\dfrac{1}{2}\times\dfrac{184}{185}=\dfrac{92}{185}\)
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
dấu " . " là dấu nhân nha bn, nãy viết nhanh quên lớp
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
A = 1.5 + 5.9 + 9.13 + ...+ 41.45
A = \(\dfrac{12}{12}\).(1.5 + 5.9 + 9.13 +...+ 41.45)
A = \(\dfrac{1}{12}\).(1.5.12 + 5.9.12 + 9.13.12 + ... + 41.45.12)
A = \(\dfrac{1}{12}\).[1.5.(9 + 3) + 5.9.(13 - 1) + 9.13.(17 - 5) + ... + 41.45.(49 - 37)]
A = \(\dfrac{1}{12}\)[1.5.9 + 1.5.3 + 5.9.13 - 1.5.9 + 9.13.17 - 5.9.13 +...+41.45.49 - 41.45.37]
A =\(\dfrac{1}{12}\).[1.5.3 + 41.45.49]
A = \(\dfrac{1}{12}\).[15 + 90405]
A = \(\dfrac{1}{12}\).90420
A = 7535
.
3/(1×5) + 3/(5×9) + 3/(9×13) + 3/(13×17) + 3/(17×21)
= 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 3/4 × (1 - 1/21)
= 3/4 × 20/21
= 5/7
Đáp án + Giải thích các bước giải:
7/1×5+7/5×9+7/9×13+7/13×17+7/17×21
=7/4×(4/1×5+4/5×9+4/9×13+4/13×17+4/17×21)
=7/4×(1−1/5+1/5−1/9+1/9−1/13+1/13−1/17+1/17−1/21)
=7/4×(1+0+0+0+0−1/21)
=7/4×(1−1/21)
=7/4×2021
=5/3
nhớ tick nha
\(\dfrac{7}{1\times5}+\dfrac{7}{5\times9}+\dfrac{7}{9\times13}+\dfrac{7}{13\times17}\)\(+\) \(\dfrac{7}{17\times21}\)
= \(\dfrac{7}{4}\times\)( \(\dfrac{4}{1\times5}\)\(+\) \(\dfrac{4}{5\times9}\)\(+\) \(\dfrac{4}{9\times13}\)\(+\)\(\dfrac{4}{13\times17}\)\(+\)\(\dfrac{4}{17\times21}\))
= 7 \(\times\)(\(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{17}-\dfrac{1}{21}\))
= 7\(\times\)(\(\dfrac{1}{1}-\dfrac{1}{21}\))
= \(\dfrac{7}{4}\)\(\times\) \(\dfrac{20}{21}\)
= \(\dfrac{5}{3}\)
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
\(B=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{125\times129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{125}-\dfrac{1}{129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{129}\right)=\dfrac{1}{4}\times\dfrac{128}{129}=\dfrac{32}{129}\)
A = 2/(1×5) + 2/(5×9) + 2/(9×13) + ... + 2/(2001×2005)
= 1/2.(1 - 1/5 + 1/5 - 1/9 + 1/9 . 1/13 + ... + 1/2001 - 1/2005)
= 1/2 . (1 - 1/2005)
= 1/2 . 2004/2005
= 1002/2005