tính nhanh 1x3+3x5+5x7+...+19x21+21x23
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có C= 1x3+2x4+3x5+...+18x20
C= 1x(1+2)+2x(1+3)+3x(1+4)+....+18x(1+19)
C= 1x1+1x2+1x2+2x3+1x3+3x4+....+1x18+18x19
C= 1+1x2+2+2x3+3+3x4+....+18+18x19
C= (1+2+3+4+...+18) + (1x2+2x3+3x4+....+18x19)
C= 171+ (1x2+2x3+....+18x19)
Đặt A=1x2+2x3+3x4+...+18x19
3A=1x2x3+2x3x3+3x4x3+....+18x19x3
3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+.....+18x19x(20-17)
3A= 1x2x3-0+2x3x4-1x2x3+3x4x5-2x3x4+.....+18x19x20-17x18x19
3A= 18x19x20
3A= 6840
A=2280
Vậy C= 171+2280=2451
A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) + 1/(9x11)
A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) + 2/(9x11)
Nhận xét :
2/(1x3) = 1 - 1/3
2/(3x5) = 1/3 - 1/5
2/(5x7) = 1/5 - 1/7
2/(7x9) = 1/7 - 1/9
2/(9x11) = 1/9 - 1/11
A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11
A x 2 = 1 - 1/11
A x 2 = 10/11
A = 10/11 : 2 = 5/11
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
Ta có : A = 1.3 + 3.5 + 5.7 + ....... + 37.39 + 39.41
=> 6A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ....... + 39.41.43
=> 6A = 39.41.43
=> A = 39.41.43 / 6
=> A = 11459,5
A=1x3+3x5+5x7+...+37x39+39x41
6A=1x3x5-1x3x5+3x5x7-3x5x7+...+39x41x43
6A=39x41x43
A=\(\frac{39x41x43}{6}\)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)