1+4+9+16+...............+9801+10000
nhanh hộ tớ nhé
nhanh đi nào
Những câu hỏi liên quan
4/3* 9/8*16/15*25/24*.....*9801/9800
(4/ 3* 9/ 8* 16/ 15* 25/ 24......9801/ 9800)
=(2* 2* 3* 3* 4* 4* 5* 5......*99* 99)/ (1* 3* 2* 4* 3* 5* 4* 6......98* 100)
=((2* 3* 4* 5*......*99)/ (1* 2* 3* 4*....* 98))/ ((2* 3* 4*...99)/ (3* 4* 5*.....*100))
=99* (1/ 50)
=99/ 50
đúng thì k cho mình bn nha (^.^)
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Tính nhanh:
A = ( 1 - 1/4 ) x ( 1 - 1/9 ) x ( 1 - 1/16 ) x ( 1 - 1/25 ) x.....x ( 1 - 1/9801 )
Giúp mik với ạ! Cảm ơn!
\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\)
\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)
\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)
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Chứng tỏ rằng: \(\dfrac{1}{2}< \dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{9801}+\dfrac{1}{10000}< \dfrac{37}{50}\)
Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)
\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)
\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)
.......
\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)
\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)
\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)
= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)
= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)
= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)
\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)
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Bài 1:Tính
M=1+9+25+49+...+9801
Bài 2:Tính
A=4+16+36+64+...+10000
Bài 3:Tính
A=1+2+4+8+...+4096+8192
1+4+9+...+9801=
1+4+9+...+9801 = (9081+1)x [(9801 - 1) : 5 + 1] : 2
= 9802 x 4901 : 2
= 24019801
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_ _ _ _ _. is bigger, the sun or the earth?
mọi người điền đầy đủ vào chỗ trống nhé
nhanh lên hộ mik
Chứng tỏ rằng: \(\frac{1}{2}< \frac{1}{2}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{9801}+\frac{1}{10000}< \frac{37}{50}\)
1+4+9+16+...+100
nhanh nhé bạn nào nhanh nhất tớ tick nhé
ta có 1^2 =1 4=2^2 cứ vậy ta đến 100=10^10 từ đó ta dễ dàng bấm máy tính từ 1^2+2^2+3^2+..+10^2 = 385
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ta có 1^2 =1 4=2^2 cứ vậy ta đến 100=10^10 từ đó ta dễ dàng bấm máy tính từ 1^2+2^2+3^2+..+10^2 = 385
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Xem thêm câu trả lời
bai 1 : a = 1.2+2.3+3.4+4.5+5.6+............+99.100
1.3+3.5+5.7 +9.11 +..........+2011.2013
1+4+9+16+ ...............+ 9801+10000
a=1.2.3+2.3.4+..........+98.99.100
bai 2 : a ) [ -2008.57+1004.(-86 )]:[32.74+16.(-48)
b) 1+2-3-4+5+6-7-8+9+10-.........................+2006-2007-2008+2009
Bài 1 : Ta có : a = 1.2 + 2.3 + 3.4 + ....... + 99.100
=> 3a = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...... + 99.100.(101 - 98)
=> 3a = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101
=> 3a = 99.100.101
=> a = \(\frac{99.100.101}{3}=333300\)
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