a, Tính tổng A = 1 + 5 + 52 + 53 + ... + 52008 + 52009
b, B = 2100 \(-\) 299 + 298 \(-\) 297 + ... + 22
c, C = ( 1000 - 13 ) . ( 1000 - 23 ) . (1000 - 33 ) ... ( 1000 - 503)
K
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Những câu hỏi liên quan
2 tháng 10 2021
a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
BD
5 tháng 8 2023
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)
Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B
\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)
Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C
\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)
\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)
HQ
Hà Quang Minh
Giáo viên
5 tháng 8 2023
\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)
DT
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 12 2020
\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+...+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)
\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)
\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)
NN
1
26 tháng 12 2023
\(2^{100}-2^{99}+2^{98}-2^{97}+2^{96}-2^{95}+...+2^4-2^3+2^2\)
\(=\left(2^{100}-2^{99}+2^{98}\right)-\left(2^{97}-2^{96}+2^{95}\right)+...+\left(2^4-2^3+2^2\right)\)
\(=2^{96}\left(2^4-2^3+2^2\right)-2^{93}\left(2^4-2^3+2^2\right)+...+\left(2^4-2^3+2^2\right)\)
\(=12\left(2^{96}-2^{93}+...+1\right)⋮12\)
CM
20 tháng 2 2019
a, Ta có :
A = 1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100
2A = 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101
A = 2A – A = ( 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 ) –( 1 + 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 )
= 2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 + 2 101 – 1 - 2 - 2 2 - 2 3 - 2 4 - . . . - 2 99 - 2 100
= 2 101 - 1
Vậy A = 2 101 - 1
b, Ta có.
B = 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99
5 2 B = 5 2 ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )
25B = 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101
25B – B = ( 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 ) – ( 5 + 5 3 + 5 5 + . . . + 5 97 + 5 99 )
24B = 5 3 + 5 5 + . . . + 5 97 + 5 99 + 5 101 – 5 - 5 3 - 5 5 - . . . - 5 97 - 5 99
24B = 5 101 - 5
B = 5 101 - 5 24 = 5 5 100 - 1 24
Vậy B = 5 5 100 - 1 24
\(A=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(5.A=5.(1+5+5^2+5^3+...+5^{2008}+5^{2009}) \)
\(5.A=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
\(5.A-A=4.A=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+5^3+...+5^{2008}+5^{2009})\)
\(4.A=5^{2010}-1\)
\(A=\frac{5^{2010}-1}{4}\)
\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2\)
\(2.B=2.(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2)\)
\(2.B=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3\)
\(2.B+B=3.B=(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3)+(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2)\)
\(3.B=2^{101}+2^2 \)
\(B=\frac{2^{101}+2^{2}}{3}\)
\(C=(1000-1^3).(1000-2^3).(1000-3^3)...(1000-50^3)\)
\(C=(1000-1^3).(1000-2^3).(1000-3^3)...(1000-10^3)...(1000-50^3)\)
\(C=(1000-1^3).(1000-2^3).(1000-3^3)...(1000-1000)...(1000-50^3)\)
\(C=(1000-1^3).(1000-2^3).(1000-3^3)...0...(1000-50^3)\)
\(C=0\)
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