tính A= 1.2+3.4+5.6+7.8+...+99.100
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Mình làm mẫu 1 bài nha !
Có : 12A = 1.5.12+5.9.12+....+101.105.12
= 1.5.12+5.9.(13-1)+.....+101.105.(109-97)
= 1.5.12+5.9.13-1.5.9+.....+101.105.109-97.101.105
= 1.5.12-1.5.9+101.105.109
= 1155960
=> A = 1155960 : 12 = 96330
Tk mk nha
Có : 4D = 1.2.3.4+2.3.4.4+....+98.99.100.4
= 1.2.3.4+2.3.4.(5-1)+.....+98.99.100.(101-97)
= 1.2.3.4+2.3.4.5-1.2.3.4+......+98.99.100.101-97.98.99.100
= 98.99.100.101
=> D = 98.99.100.101/4 = 24497550
3F = 1 . 2 . 3 + 3 . 4 . ( 5 - 2 ) + 5 . 6 . ( 7 - 4 ) +.....+ 99 . 100 . (101 - 98 )
3F = 1. 2 . 3 + 3. 4 . 5 - 2 . 3 . 4 + 5 . 6 . 7 - 4 . 5 . 6 +.....+ 99 . 100 . 101 - 98 . 99 . 100
3F = 1 . 2 . 3 + 99 . 100. 101
3F = 3 . 2 + 3 . 33 . 100 . 101
3F = 3 ( 2 + 333 300)
=>F = 3 . 333 302 : 3
=> F = 333 302
Vậy F = 333 302
TRả lời ;..............................................
SCSH: ( 100 - 1,2 ) : 2 = 49,4
Tổng: ( 100 + 1,2 ) : 2 = 50,6
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
3A = 1.2.3+2.3(4-1)+3.4.(5-2)+.+99.100.(101-98)
3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.+99.100.101-98.99.100
3A = 99.100.101
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\left(đpcm\right)\)
Ta có : \(VT=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
\(\Rightarrow\) \(ĐPCM\)
A=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=1/3-1/9
A=2/9
các câu 2;3 còn lại giống câu 1 bạn nhé
bạn thay số vào rồi làm tương tự
\(A=\left(2-1\right).2+\left(4-1\right).4+\left(6-1\right).6+...+\left(100-1\right).100\\ A=2^2-2+4^2-4+6^2-6+...+100^2-100\\ A=\left(2^2+4^2+...+100^2\right)-\left(2+4+...+100\right)\\ A=2^2\left(1+2^2+3^2+...+50^2\right)-\dfrac{\left(100+2\right).50}{2}\\ A=\dfrac{4.50.51.52}{6}-\dfrac{102.50}{2}=85850\)
Ta có: A = (2 – 1).2 + (4 – 1).4 + (6 – 1).6 + … + (100 – 1).100
A = 22 – 2 + 42 – 4 + 62 – 6 + … + 1002 – 100
A = (22 + 42 + 62 + … + 1002) – (2 + 4 + 6 + … + 100)
A = 22.(12 + 22 + 32 + … + 502) – (100 + 2).50 : 2
A = 22.50.51.52 : 6 – 51.50 = 88400 – 2550 = 85850.
giúp mình với
cảm ơn !
C1:
\(A=1.2+2.3+3.4+4.5+....+99.100\\ \Rightarrow3A=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+....+99.100\left(101-98\right)\\ \Rightarrow3A=1.2.3-0.1.2+2.3.4-1.2.3+...+99.100.101-98.99.100\\ \Rightarrow3.A=99.100.101\\ \Rightarrow A=\frac{99.100.101}{3}=333300\)