tính giá trị biểu thức sau:B=1+3+5+7+...+97+99
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.
b)
B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 99-97+95-93+...+7-5+3-1
A = 2 + 2 +...+ 2 + 2 ( có 100 số )
A = 100 : 2
A = 50
A=99-97+95-93+......+7-5+3-1
A=2+2+.........+2+2(CÓ 50 CẶP)
A=2.50
A=50.2
A=100
![](https://rs.olm.vn/images/avt/0.png?1311)
D = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{4}{3x5}+\dfrac{4}{5x7}+\dfrac{4}{7x9}+...+\dfrac{4}{97x99}+\dfrac{4}{99x101}\)
\(A=4x\left(\dfrac{1}{3x5}+\dfrac{1}{5x7}+\dfrac{1}{7x9}+...+\dfrac{1}{97x99}+\dfrac{1}{99x101}\right)\)
\(A=4x\left[\dfrac{1}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\dfrac{1}{2}x\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\dfrac{1}{2}x\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\dfrac{1}{2}x\left(\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}x\left(\dfrac{1}{99}-\dfrac{1}{101}\right)\right]\)
\(A=4x\dfrac{1}{2}x\left[\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right]\)
\(A=2x\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2x\dfrac{98}{303}=\dfrac{916}{303}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 1+(2-3)+(5-4)+...+(98-99)-100
=1-1+1-1+...+1-1-100
=-100
![](https://rs.olm.vn/images/avt/0.png?1311)
Số các số hạng của \(B\) là:
\(\left(99-1\right):2+1=50\left(số\right)\)
Tổng \(B\) bằng:
\(\left(99+1\right)\cdot50:2=2500\)
Vậy \(B=2500\)
B = 1+3+5+7+...+97+99
Tổng B có số số hạng là: (99 - 1) : 2 +1 = 50(số hạng)
B = (99 + 1) . 50 : 2 = 100 . 50 :2 = 2500