a)2/1x3+2/3x5+2/5x7+...+2/37x99
b)3/1x4+3/+3/4x7+...+3/97x100
c)1x3+3x5+...+47x49
d)1x2x3+2x3x4+...+18x19x20
2)Không tính cụ thể hãy so sánh
a)2013x2017và2015x2015
b)2013x2017và2014x2016
Em cần giải gấp các anh chị giúp em nha
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a=511/256
b=647/20
c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời
a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)
= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256
= 2 - 1/256
= 511/256
Câu b bạn có viết sai đề không vậy?
A = 3/1.3 + 3/3.5 + 3/5.7 + 3/7.9 + ... + 3/97.99
A = 3/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + .... + 2/97 - 2/99
A = 3/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 )
A = 3/2 . ( 1 - 1/99 )
A = 3/2 . 98/99
A = 49/33
b) dãy số không có quy luật==> bạn xem lại đề
c) \(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{50\times51}-\frac{1}{51\times52}\right)\)
\(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{51\times52}\right)=\frac{1}{2}\times\frac{2650}{5408}=\frac{1325}{5408}\)
Tính nhanh
a) 5/1x3 + 5/3x5 + 5/5x7 + ........ + 5/43x45
b) 6/1x4 + 6/4x7 + 6/7x10 + ...... + 6/97x100
\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
Lời giải:
$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}$
$=1-\frac{1}{99}=\frac{98}{99}$
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)