A=2+22 +23+24+...+2100
B=1+5+52+53+...+5150
C=3+32+33+...+31000
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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) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
CM
21 tháng 12 2019
a, Ta có:
2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100
= 2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100
= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4
= 2 . 31 + 2 6 . 31 + . . . + 2 96 . 31
= 2 + 2 6 + . . . + 2 96 . 31 chia hết cho 31
b, Ta có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5
= 5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6
= ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6 chia hết cho 6
Ta lại có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150 (có đúng 25 nhóm)
= [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... + [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]
= [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... + [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]
= ( 5 . 126 + 5 2 . 126 + 5 3 . 126 ) + ... + ( 5 145 . 126 + 5 146 . 126 + 5 147 . 126 )
= ( 5 + 5 2 + 5 3 ) . 126 + ( 5 7 + 5 8 + 5 9 ) . 126 + ... + ( 5 145 + 5 146 + 5 147 ) . 126
= 126.[ ( 5 + 5 2 + 5 3 ) + ( 5 7 + 5 8 + 5 9 ) + ... + ( 5 145 + 5 146 + 5 147 ) ] chia hết cho 126.
Vậy 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150 vừa chia hết cho 6, vừa chia hết cho 126
14 tháng 10 2023
\(A=2+2^2+...+2^{20}\)
\(2A=2^2+2^3+...+2^{21}\)
\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)
\(A=2^{21}-2\)
___________
\(B=5+5^2+...+5^{50}\)
\(5B=5^2+5^3+...+5^{51}\)
\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)
\(4B=5^{51}-5\)
\(B=\dfrac{5^{51}-5}{4}\)
___________
\(C=1+3+3^2+...+3^{100}\)
\(3C=3+3^2+...+3^{101}\)
\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)
\(2C=3^{101}-1\)
\(C=\dfrac{3^{101}-1}{2}\)
2 tháng 9 2021
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
30 tháng 12 2024
a: 23−53:52+12⋅2223−53:52+12⋅22
=8−5+48=8−5+48
=51=51
b: 5⋅[(85−35:7):8+90]−55⋅[(85−35:7):8+90]−5
=5⋅[10+90]−5=5⋅[10+90]−5
=495
C
22 tháng 12 2023
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
a.
$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$
$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$
$\Rightarrow S=2^{2018}-1$
b.
$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$
$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$
$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu c, d bạn làm tương tự a,b.
c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$
d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$
10 tháng 10 2023
a) \(S=1+2+2^2+..+2^{2022}\)
\(2S=2+2^2+2^3+...+2^{2023}\)
\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)
\(S=2^{2023}-1\)
b) \(S=3+3^2+3^3+...+3^{2022}\)
\(3S=3^2+3^3+...+3^{2023}\)
\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)
\(2S=3^{2023}-3\)
\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)
c) \(S=4+4^2+4^3+...+4^{2022}\)
\(4S=4^2+4^3+...+4^{2023}\)
\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)
\(3S=4^{2023}-4\)
\(S=\dfrac{4^{2023}-4}{3}\)
d) \(S=5+5^2+...+5^{2022}\)
\(5S=5^2+5^3+...+5^{2023}\)
\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)
\(4S=5^{2023}-5\)
\(S=\dfrac{5^{2023}-5}{4}\)
`A = 2 + 2^2+ 2^3 + 2^4 + ... + 2^100`
`2A = 2 .(2 + 2^2+ 2^3 + 2^4 + ... + 2^100)`
`2A = 2^2 + 2^3 + 2^4 + 2^5 + ... + 2^101`
`2A - A= (2^2 + 2^3 + 2^4 + 2^5 +...+ 2^101)-(2 + 2^2+ 2^3 + 2^4 + ... + 2^100)`
`A= 2^101 - 2`
`B = 1 + 5 + 5^2 + 5^3 + ... + 5^150`
`5B = 5 . (1 + 5 + 5^2 + 5^3 + ... + 5^150)`
`5B = 5 + 5^2 + 5^3 + 5^4 + ... + 5^151`
`5B - B = (5 + 5^2 + 5^3 + 5^4 + ... + 5^151)-(1 + 5 + 5^2 + 5^3 + ... + 5^150)`
`4B= 5^151 - 1`
`B = (5^151 - 1)/4`
`C = 3 + 3^2+3^3 + ... + 3^1000`
`3C = 3. (3 + 3^2+3^3 + ... + 3^1000)`
`3C = 3^2 + 3^3 + 3^4 + ... + 3^1001`
`3C-C = (3^2 + 3^3 + 3^4 + ... + 3^1001)-(3 + 3^2+3^3 + ... + 3^1000)`
`2C = 3^1001 - 3`
`C = (3^1001 - 3)/2`
\(A=2+2^2+2^3+...+2^{100}\)
=>\(2A=2^2+2^3+...+2^{101}\)
=>\(2A-A=2^2+2^3+...+2^{101}-2-2^2-...-2^{100}\)
=>\(A=2^{101}-2\)
\(B=1+5+5^2+...+5^{150}\)
=>\(5B=5+5^2+5^3+...+5^{151}\)
=>\(5B-B=5+5^2+...+5^{151}-1-5-...-5^{150}\)
=>\(4B=5^{151}-1\)
=>\(B=\dfrac{5^{151}-1}{4}\)
\(C=3+3^2+...+3^{1000}\)
=>\(3C=3^2+3^3+...+3^{1001}\)
=>\(3C-C=3^2+3^3+...+3^{1001}-3-3^2-...-3^{1000}\)
=>\(2C=3^{1001}-3\)
=>\(C=\dfrac{3^{1001}-3}{2}\)