a=2mu 101 - 2

b= 3 mu 2010 - 1

c=5mu 1999-1

d=4 mu n . 4 -4

a=2+2²+...+2¹⁰⁰

2a=2²+2³+2⁴+...+2¹⁰¹

a=2a-a=a

=> a= 2²+2³+2⁴+..+2¹⁰¹ -(2+2^2+...+2^100)

=>a= 2^101 -2 

sai

sai

đúng

sai

\(37\left(3+7\right)=3^3+7^3\)

\(37\cdot10=27+343\)

\(370=370\)(thảo mãn

=>đúng

vậy 37(3+7)=3^3+7^3

1. 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101 
=> 3A - A = (3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 ) 
=> 2A = 3^101 - 3 => 2A + 3 = 3^101 vậy n = 101 
2. 2A = 8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21 
=> 2A - A = (8 + 2 ^ 3 + 2^4 + ... + 2^20 + 2^21) - (4+ 2^2 + 2 ^ 3 + 2^4 + ... + 2^20 ) 
=> A = 2^21 là một lũy thừa của 2 
3. 
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101 
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100) 
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2 
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101 
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 ) 
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2 
c) xem lại đề ý c xem quy luật như thế nào nhé. 
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151 
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150) 
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2

help

Đề ghi sai rồi em

\(D=6+6^1+6^2+6^3+\cdots+6^{120}\) ko chia hết cho 7 và 43

Mà \(D=6^1+6^2+6^3+\cdots+6^{120}\) mới đúng

Em ghi thừa số 6 ở đầu thì phải

1: \(A=2+2^2+2^3+\cdots+2^{100}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{101}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{101}-2-2^2-2^3-\cdots-2^{100}\)

=>\(A=2^{101}-2\)

2: \(B=1+5+5^2+5^3+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

3: \(C=3+3^2+\cdots+3^{1000}\)

=>\(3C=3^2+3^3+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

Câu 1:

A = 2 + 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)

2A = 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\) + 2\(^{101}\)

2A - A = (2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)+ 2\(^{101}\)) -(2 + 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\))

A = 2\(^2\) + 2\(^3\) + ... + 2\(^{100}\)+ 2\(^{101}\) - 2 - 2\(^2\) -2\(^3\) - ... - 2\(^{100}\)

A = (2\(^2\) - 2\(^2\)) + (2\(^3\) - 2\(^3\)) + ... + (2\(^{100}\) - 2\(^{100}\)) + (2\(^{101}\) - 2)

A = 0 + 0 + 0 + ... + 0 + 2\(^{101}\) - 2

A = 2\(^{101}\) - 2

a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2

a) Có A=\(1+3+3^2+3^3+....+3^{100}\)

\(\Rightarrow\)3A =\(3\left(1+3+3^2+3^3+...+3^{100}\right)\)=\(3+3^2+3^3+3^4+...+3^{101}\)

\(\Rightarrow2A=3+3^2+3^3+....+3^{101}-1-3-3^2-3^3-....-3^{100}=3^{101}-1\)\(\Rightarrow A=\dfrac{3^{101}-1}{2}\)

Bài b/c/d : bn cứ lm tương tự.

Ngu xi 

Giải:

a) \(4^n:4=64\)

\(\Leftrightarrow4^{n-1}=64\)

\(\Leftrightarrow4^{n-1}=4^3\)

Vì \(4=4\)

Nên \(n-1=3\)

\(\Leftrightarrow n=4\)

b) \(7^5:7^n=49\)

\(\Leftrightarrow7^{5-n}=49\)

\(\Leftrightarrow7^{5-n}=7^2\)

Vì \(7=7\)

Nên \(5-n=2\)

\(\Leftrightarrow n=3\)

c) \(3^n=27\)

\(\Leftrightarrow3^n=3^3\)

Vì \(3=3\)

Nên \(n=3\)

d) \(11^n=121\)

\(\Leftrightarrow11^n=11^2\)

Vì \(11=11\)

Nên \(n=2\)

e) \(5.5^n=125\)

\(\Leftrightarrow5^{1+n}=125\)

\(\Leftrightarrow5^{1+n}=5^3\)

Vì \(5=5\)

Nên \(1+n=3\)

\(\Leftrightarrow n=2\)

g) \(4^n=64:4\)

\(\Leftrightarrow4^n=16\)

\(\Leftrightarrow4^n=4^2\)

Vì \(4=4\)

Nên \(n=2\)

Chúc bạn học tốt!

a) \(4^n\div4=64\)

\(\Rightarrow4^n=64\div4\)

\(\Rightarrow4^n=16\)

\(\Rightarrow4^n=4^2\)

\(\Rightarrow\) n = 2

b) \(7^5\div7^n=49\)

\(\Rightarrow7^5\div7^n=7^2\)

\(\Rightarrow7^n=7^5\div7^2\)

\(\Rightarrow7^n=7^3\)

\(\Rightarrow\) n = 3

c) \(3^n=27\)

\(\Rightarrow3^n=3^3\)

\(\Rightarrow\) n = 3

d) \(11^n=121\)

\(\Rightarrow11^n=11^2\)

\(\Rightarrow\) n = 2

e) \(5\times5^n=125\)

\(\Rightarrow5^n=125\div5\)

\(\Rightarrow5^n=25\)

\(\Rightarrow5^n=5^2\)

\(\Rightarrow\) n = 2

g) \(4^n=64\div4\)

\(\Rightarrow4^n=16\)

\(\Rightarrow4^n=4^2\)

\(\Rightarrow\) n = 2