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S=1x3+3x5+5x7+...+49x51
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NQ
2
NH
1 tháng 2 2020
\(S=\frac{4}{1\times3}+\frac{16}{3\times5}+\frac{36}{5\times7}+...+\frac{2500}{49\times51}\)
\(=\frac{1\times3+1}{1\times3}+\frac{3\times5+1}{3\times5}+\frac{5\times7+1}{5\times7}+...+\frac{49\times51+1}{49\times51}\)
\(=\frac{1\times3}{1\times3}+\frac{1}{1\times3}+\frac{3\times5}{3\times5}+\frac{1}{3\times5}+\frac{5\times7}{5\times7}+\frac{1}{5\times7}+...+\frac{49\times51}{49\times51}+\frac{1}{49\times51}\)
\(=1+\frac{1}{1\times3}+1+\frac{1}{3\times5}+1+\frac{1}{5\times7}+...+\frac{1}{49\times51}\) ( Có : \(\left(51-3\right)\div2+1=25\)chữ số 1 )
\(=25+\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{49\times51}\)
\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{49}-\frac{1}{51}\right)\)
\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=25+\frac{1}{2}\times\left(1-\frac{1}{51}\right)\)
\(=25+\frac{1}{2}\times\frac{50}{51}\)
\(=25+\frac{25}{51}\)
\(=\frac{1300}{51}\)
1 tháng 2 2020
\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)
\(=\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+...+\frac{2500}{2499}\)
\(=1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+...+1+\frac{1}{2499}\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2500}\right)\)
\(=25+\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\right)\)
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}=\frac{50}{51}\)
\(\Rightarrow S=25+\frac{50}{51}=\frac{1325}{51}\)
Vậy S=\(\frac{1325}{51}\)
LM
1
DD
6 tháng 8 2015
\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
S
0
S
0
N
1
DP
6 tháng 7 2017
Đặt \(S=\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{49\cdot51}\)
\(S=\frac{3}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(S=\frac{3}{2}\cdot\left(1-\frac{1}{51}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\frac{50}{51}=\frac{3\cdot50}{2\cdot51}=\frac{150}{102}=\frac{25}{17}\)
24 tháng 3 2019
Ta có:
\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+........+\frac{2500}{49.51}\)
CD
6
PL
2 tháng 4 2019
Dễ thôi bạn à
\(A=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)
\(A=\frac{1.3+1}{1.3}+\frac{3.5+1}{3.5}+\frac{5.7+1}{5.7}+...+\frac{49.50+1}{49.51}\)
\(A=\frac{1.3}{1.3}+\frac{1}{1.3}+\frac{3.5}{3.5}+\frac{1}{3.5}+\frac{5.7}{5.7}+\frac{1}{5.7}+...+\frac{49.51}{49.51}+\frac{1}{49.51}\)
\(A=1+\frac{1}{1.3}+1+\frac{1}{3.5}+1+\frac{1}{5.7}+...+1+\frac{1}{49.51}\) (có: (51 - 3) : 2 + 1 = 25 chữ số 1)
\(A=25+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\frac{50}{51}\)
\(A=25+\frac{25}{51}\)
\(A=\frac{1300}{51}\)
HH
0
18 tháng 3 2019
A=1/6+1/12+1/20+1/30+1/42+1/56+1/72
A=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9
A=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
A=1/2-1/9
Câu B tương tự nha bạn :333
\(S=1\cdot3+3\cdot5+...+49\cdot51\)
\(=1\left(1+2\right)+3\left(3+2\right)+...+49\left(1+2\right)\)
\(=\left(1^2+3^2+...+49^2\right)+2\left(1+3+...+49\right)\)
\(=\left[1^2+3^2+...+\left(2\cdot25-1\right)^2\right]+2\cdot\left[\left(\dfrac{49-1}{2}+1\right)\cdot\dfrac{\left(49+1\right)}{2}\right]\)
\(=\dfrac{25\left(4\cdot25^2-1\right)}{3}+2\cdot25\cdot\dfrac{50}{2}\)
\(=25\cdot\dfrac{\left(4\cdot625-1\right)}{3}+25\cdot50\)
\(=25\cdot833+25\cdot50=22075\)
Chúng ta hãy tìm tổng của dãy số S=1⋅3+3⋅5+5⋅7+…+49⋅51S = 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + 49 \cdot 51.
Dãy này có thể được viết lại dưới dạng:
S=∑k=125(2k−1)(2k+1)S = \sum_{k=1}^{25} (2k-1)(2k+1)Chúng ta có thể nhận thấy rằng:
(2k−1)(2k+1)=(2k)2−1(2k-1)(2k+1) = (2k)^2 - 1Vậy tổng SS có thể được viết lại như sau:
S=∑k=125[(2k)2−1]S = \sum_{k=1}^{25} [(2k)^2 - 1] =∑k=125(4k2−1)= \sum_{k=1}^{25} (4k^2 - 1) =4∑k=125k2−∑k=1251= 4 \sum_{k=1}^{25} k^2 - \sum_{k=1}^{25} 1Ta biết rằng:
∑k=125k2=25⋅26⋅516=5525\sum_{k=1}^{25} k^2 = \frac{25 \cdot 26 \cdot 51}{6} = 5525 ∑k=1251=25\sum_{k=1}^{25} 1 = 25Do đó:
S=4⋅5525−25=22100−25=22075S = 4 \cdot 5525 - 25 = 22100 - 25 = 22075Vậy, tổng của dãy số là 2207522075. Khó lắm đấy