tính A=1x3+2x4+3x5+4x6+...+48x50+49x51
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Những câu hỏi liên quan
tính tổng S = 1x3+2x4+3x5+...+48x50+49x50
tính : 1x3 + 2x4 + 3x5 + 4x6 + ... + 99x101 . Tính giúp mik nhé :))))
khó dữ vậy ba ?????
1x3+2x4+3x5+4x6+.......+99x101+100x102
=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)
=1.2+1+2.3+2+3.4+3+...+100.101+100
=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)
=333300+5000
=338300
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1x3+2x4+3x5+4x6+...+90x92
\(S=\frac{1}{1x3}-\frac{1}{2x4}+\frac{1}{3x5}-\frac{1}{4x6}+\frac{1}{5x7}-\frac{1}{6x8}+\frac{1}{7x9}-\frac{1}{8x10}\)
giúp mình với mình đang cần gấp.
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
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S=1/1x3-1/2x4-1/3x5-1/4x6+1/5x7
\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+\dfrac{1}{5.7}\)
\(S=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}\)
\(S=1+\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{31}{21}\)
Chúc bạn học tốt!!!
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Chứng minh : A = 1/1x3 + 1/2x4 + 1/3x5 + 1/4x6 + ... + 1/97x99 + 1/98x100 < 3/4
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
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(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x(1+1/4x6)x(1+1/5x7)
A= 22+42+62+...+2002
B= 1x3+3x5+5x7+...+99x101
C= 2x4+4x6+6x8+...+98x100x102
D= 1x2+3x4+5x6+...+99x100
Mình đang vội các bạn giúp mình với!