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a: \(\dfrac{1}{2}+\dfrac{-3}{8}+\dfrac{5}{9}\)
\(=\dfrac{36}{72}-\dfrac{27}{72}+\dfrac{40}{72}\)
\(=\dfrac{49}{72}\)
b: \(\dfrac{13}{-30}+\dfrac{17}{45}+\dfrac{-7}{18}\)
\(=\dfrac{-13}{30}+\dfrac{17}{45}+\dfrac{-7}{18}\)
\(=\dfrac{-39+34-35}{90}=\dfrac{-40}{90}=-\dfrac{4}{9}\)
c: \(-\dfrac{5}{19}+\dfrac{7}{4}-\left(-\dfrac{2}{3}\right)+\dfrac{5}{6}\)
\(=-\dfrac{5}{19}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{5}{6}\)
\(=\dfrac{-5}{19}+\dfrac{21+8+10}{12}\)
\(=-\dfrac{5}{19}+\dfrac{39}{12}=\dfrac{-5}{19}+\dfrac{13}{4}=\dfrac{227}{76}\)
Số số hạng trong tổng trên là:
( n -1 ) : 1 + 1 = n -1
Tổng trên có giá trị là
( n + 1 ) . ( n - 1 ) : 2 = ( n2 - 1 ) : 2
Thông cảm nha >
Mình giải nhầm
Đáp án đúng ở câu hỏi của cậu sau ấy
Ta có : a) A= 1+ 5+ 52+ 53+........+ 51998
=> 5A = 5+ 52+ 53+........+ 51999
=> 5A - A = 51999 - 1
=> 4A = 51999 - 1
\(\Rightarrow A=\frac{5^{1999}-1}{4}\)
b) Ta có : b) B= 1+ 4+ 42 + ...... + 4n
=> 4B = 4 + 42 + 43 + ...... + 4n + 1
=> 4B - B = 4n + 1 - 1
=> 3B = 4n + 1 - 1
=> \(B=\frac{4^{n+1}-1}{3}\)
a) 1+2+3+4+5+...+n = n(n+1) / 2
b)2+4+6+...+2n = [(2n-2):2+1] . (2n+2)/2 = n . ( 2n+2) /2
a) \(1+2+3+4+...+n\)
\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right):2\)
\(=n\left(n+1\right):2\)
\(=\dfrac{n\left(n+1\right)}{2}\)
b) \(2+4+6+..+2n\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
c) \(1+3+5+...+\left(2n+1\right)\)
\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)
\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)
\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)
\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)
\(=\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n+1\right)\)
d) \(1+4+7+10+...+2005\)
\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)
\(=2006\cdot\left(2004:3+1\right):2\)
\(=2006\cdot\left(668+1\right):2\)
\(=1003\cdot669\)
\(=671007\)
e) \(2+5+8+...+2006\)
\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)
\(=2008\cdot\left(2004:3+1\right):2\)
\(=1004\cdot\left(668+1\right)\)
\(=1004\cdot669\)
\(=671676\)
g) \(1+5+9+...+2001\)
\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)
\(=2002\cdot\left(2000:4+1\right):2\)
\(=1001\cdot\left(500+1\right)\)
\(=1001\cdot501\)
\(=501501\)
Cái tên.. àk mà thôi -_-
\(a)\) \(1+2+3+4+...+n=\frac{n\left(n+1\right)}{2}\)
\(b)\) \(2+4+6+8+...+2n=\left(\frac{2n-2}{2}+1\right)\left(2n+2\right)=\frac{2n\left(2n+2\right)}{2}=2n\left(n+1\right)\)
\(c)\) \(1+3+5+...+\left(2n+1\right)=\left(\frac{2n+1-1}{2}+1\right)\left(2n+1+1\right)=\frac{\left(2n+2\right)\left(2n+2\right)}{2}=\frac{\left(2n+2\right)^2}{2}\)
\(d)\) \(1+4+7+10+...+2005=\left(\frac{2005-1}{3}+1\right)\left(2005+1\right)=1342014\)
\(e)\) \(2+5+...+2006=\left(\frac{2006-2}{3}+1\right)\left(2006+2\right)=1343352\)
\(g)\) \(1+5+9+...+2001=\left(\frac{2001-1}{4}+1\right)\left(2001+1\right)=1003002\)
Chúc bạn học tốt ~
a=[(n+1)*(n-1)]/2 s để mình làm nốt cho