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\(\left(1+2^2+2^4+....+2^{100}\right)-\left(2^1+2^3+.....+2^{101}\right)\)
\(\Rightarrow A=1-2^{101}\)
Vậy \(A=1-2^{101}\)
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\(A=2^1+2^2+2^3+...+2^{99}+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{100}+2^{101}\right)-\left(2^1+2^2+2^3+...+2^{99}+2^{100}\right)\)
\(A=2^{101}-2\)
Ta có: A=2^1+2^2+2^3+...+2^99+2^100
=> 2A=2^2+2^3+2^4+...+2^100+2^101 Mà A=2^1+2^2+2^3+.........+2^100
=>2A-A=2^101-2^1
=>1A=2^101-2^1
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Ta có : (x - 6)3 = (x - 6)2
=> (x - 6)3 - (x - 6)2 = 0
=> (x - 6)2(x - 6 - 1) = 0
=> (x - 6)2 (x - 7) = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-6\right)^2=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
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Vì câu a có dấu x mk ko hiểu nên mk làm câu b nhé:
có 12 - 22 = -3
32 - 42 = -7
...................
992 - 1002 = -199
vậy chúng cách nhau 4 đơn vị
⇒ -((199 + 3).((199 - 3):4 + 1):2))) = -5050 vậy A = -5050
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ta có: 1^2 - 2^2 +3^2 -4^2 +...........+99^2-100^2+101^2
= (1-2)(1+2) + (3-4)(3+4) + (5-6)(5+6) + ....+ (99-100)(99+100) +101^2
= -3 - 7 - 11 - ....-199 + 101^2
= 101^2 - (3 + 7 + 11 + ... + 199)
[ Ta dễ thấy (3 + 7 + 11 + ... + 199) là một cấp số cộng có d=4 và n=50]
= 101^2 - [(199 + 3).50]/2
= 5151
ta có: 1^2 - 2^2 +3^2 -4^2 +...........+99^2-100^2+101^2
= (1-2)(1+2) + (3-4)(3+4) + (5-6)(5+6) + ....+ (99-100)(99+100) +101^2
= -3 - 7 - 11 - ....-199 + 101^2
= 101^2 - (3 + 7 + 11 + ... + 199)
[ Ta dễ thấy (3 + 7 + 11 + ... + 199) là một cấp số cộng có d=4 và n=50]
= 101^2 - [(199 + 3).50]/2
= 5151
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A = 2100 - 299 - 298 - ...-2-1
=> 2A = 2101 - 2100 - 299-...-22 - 2
=> 2A-A = 2101 - 2100 - 2100 + 1
A = 2101 - 2100.(1+1) + 1
A = 2101 - 2100. 2+1
A = 2101- 2101+1
A = 1
b) B = 1 - 5 + 52 - 53+...+598-599
=> 5B = 5 - 52+53-54+...+599-5100
=> 5B+B = -5100+1
6B = -5100+1
\(B=\frac{-5^{100}+1}{6}\)
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a) S=(1-2)^2+(3-4)^3+......+(99-100)^99
=(-1)^2+(-1)^3+......+(-1)^99
=1+(-1)+....+(-1)
=[1+(-1)]+[1+(-1)]+.......+[1+(-1)]
=0+0+.....+0=0
1^2-2^2+3^2-4^2+.......+99^2-100^2
=(1+2)(-1)+(3+4)(-1)+......+(99+100)(-1)
=(-1)(1+2+3+4+......+99+100)=(-1).101.100:2=-5050
A = 1.1+2.2+3.3+...+99.99
=>A= 1.(2-1)+2.(3-1)+3.(4-1)+...+99.(100-1)
=>A=1.2-1+2.3-2+3.4-3+...+99.100-99
=>A=(1.2+2.3+3.4+......+99.100)-(1+2+3+...+99)
Đặt S =1.2+2.3+3.4+...+99.100
=>3S= 1.2.3+2.3.3+3.4.3+....+99.100.3
=>3S= 1.2.3+2.3.(4-1)+3.4.(5-2)+.....+99.100.(101-98)
=>3S =1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+99.100.101-98.99.100
=>3S= 99.100.101=999900
=>S= 999900:3=333300
=>A =333300 -(99+1).99:2
=>A=333300-4950
=>A=328350
A= 12 + 22 + 32 + ... + 992 + 1002
A = 1 + 2(1 + 1) + 3(2 + 1) +... + 99(98 + 1) + 100(99 + 1)
A = 1 + 1.2 + 2 + 2.3 + 3 +... + 98.99 + 99 + 99.100 + 100
A = (1.2 + 2.3 + 3.4 +... + 99.100) + (1 + 2 + 3 +... + 99 + 100)
A = 333300 + 5050
A = 338050
Vậy A = 338050