d) D = 1.99 + 3.97 + 5.95 + ... + 99.1.
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LB
0
P
20 tháng 8 2023
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{97}+\dfrac{1}{3}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{\left(\dfrac{1}{1\cdot99}+\dfrac{1}{99\cdot1}\right)+\left(\dfrac{1}{97\cdot3}+\dfrac{1}{97\cdot3}\right)+...+\left(\dfrac{1}{51\cdot49}+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{97\cdot3}+...+\dfrac{100}{49\cdot51}}{\dfrac{2}{1\cdot99}+\dfrac{2}{97\cdot3}+...+\dfrac{2}{51\cdot49}}\)
\(=\dfrac{100\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}{2\cdot\left(\dfrac{1}{99}+\dfrac{1}{97\cdot3}+...+\dfrac{1}{49\cdot51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
C
3 tháng 2 2019
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
VL
0
Ta có: \(D=1\cdot99+3\cdot97+\cdots+99\cdot1\)
\(=2\left(1\cdot99+3\cdot97+\cdots+49\cdot51\right)\)
\(=2\left\lbrack1\left(100-1\right)+3\left(100-3\right)+\cdots+49\left(100-49\right)\right\rbrack\)
\(=2\cdot\left\lbrack100\left(1+3+\cdots+49\right)-\left(1^2+3^2+\cdots+49^2\right)\right\rbrack\)
Đặt \(A=1+3+\cdots+49\)
Số số hạng của dãy số là: \(\frac{49-1}{2}+1=\frac{48}{2}+1=24+1=25\) (số)
Tổng của dãy số là: \(\left(49+1\right)\cdot\frac{25}{2}=50\cdot\frac{25}{2}=25\cdot25=625\)
Đặt \(B=1^2+3^2+\cdots+49^2\)
\(=1^2+2^2+3^2+4^2+\cdots+50^2-\left(2^2+4^2+\cdots+50^2\right)\)
\(=\frac{50\left(50+1\right)\left(2\cdot50+1\right)}{6}-2^2\left(1^2+2^2+\cdots+25^2\right)\)
\(=\frac{50\cdot51\cdot101}{6}-4\cdot\frac{25\left(25+1\right)\left(2\cdot25+1\right)}{6}\)
\(=25\cdot17\cdot101-\frac23\cdot25\cdot26\cdot51\)
\(=25\cdot17\cdot101-25\cdot26\cdot34=25\cdot17\cdot\left(101-26\cdot2\right)\)
\(=425\cdot\left(101-52\right)=425\cdot49=20825\)