A = 4/1x3+16/3x5+36/5x7+...+2500/49x51
tính tổng của A
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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=\(\frac{1325}{51}\)
24 tháng 3 2019
Ta có:
\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+........+\frac{2500}{49.51}\)
DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
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: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
LT
2
VD
13 tháng 3 2022
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 3 2023
A = \(\dfrac{4}{1\times3}\) - \(\dfrac{8}{3\times5}\) + \(\dfrac{12}{5\times7}\) - \(\dfrac{16}{7\times9}\) + \(\dfrac{20}{9\times11}\) - \(\dfrac{24}{11\times13}\)
A = ( \(\dfrac{1}{1}+\dfrac{1}{3}\)) - ( \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\)+ \(\dfrac{1}{7}\)) - ( \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\)) +( \(\dfrac{1}{9}\)+ \(\dfrac{1}{11}\)) - (\(\dfrac{1}{11}\)+\(\dfrac{1}{13}\))
A = \(\dfrac{1}{1}+\dfrac{1}{3}\) - \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{12}{13}\)
NH
0
DT
2
BG
10 tháng 3 2018
Ta có:
A = \(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
= \(\frac{1}{1}-\frac{1}{11}\)
=\(\frac{10}{11}\)
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)
\(A=25+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\left(1-\frac{1}{51}\right)\)
\(A=25+\frac{1}{2}.\frac{50}{51}\)
\(A=25+\frac{25}{51}\)
\(A=\frac{1300}{51}\)
thank you