a) 3 ^ 6 : 3 ^ 2 + 2 ^ 3 + 2 ^ 2 

 = 3^{6 - 2}  + 2^{3 + 2} 

  = 3⁴ + 2⁵  = 81 + 32 

  = 113 .

b) ( 2100 - 63 ) : 21

 = 2037 : 21

 = 97 .

c) 32 x 59 + 41 x 32

 = 32 x ( 59 + 41 )

 = 32 x 100

 = 3200 .

d) 99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5 + 3 - 1 ( 50 so = 25 hieu )

= 2 + 2 + 2 + ... + 2 + 2 = 2 x 25 

= 50 .

a)   \(3^6:3^2+2^3.2^2=3^4+2^5=113\)

b)   \(\left(2100-63\right):21=2037:21=97\)

c)    \(32.59+41.32=32.\left(59+41\right)=32.100=3200\)

d)   \(99-97+95-93+........+9-7+5-3+1=2+2+...+2+1=25.2+1=50+1=51\)

giải giúp mik bài này nha

12x-33=3^2.3^3

\(A=4+2^2+2^3+...+2^{99}\)

=>  \(2A=8+2^3+2^4+...+2^{100}\)

=>  \(2A-A=\left(8+2^3+2^4+...+2^{100}\right)-\left(4+2^2+2^3+...+2^{99}\right)\)

=>  \(A=2^{100}< 2^{200}=2^{2.100}=4^{100}=B\)

Vậy  A < B

a) Số lượng số hạng:

\(\left(100-2\right):2+1=50\) (số hạng)

Tổng là:

\(A=\left(100+2\right)\cdot50:2=2550\) 

b) \(B=99-97+95-93+...+3-1\)

\(B=\left(99-97\right)+\left(95-93\right)+...+\left(3-1\right)\)

\(B=2+2+...+2\) (50 số hạng)

\(B=2\cdot50\)

\(B=100\)

\(A=2+4+6+...+100\)

\(A=\left[\left(100-2\right):2+1\right].\left(100+2\right)=50.102=5100\)

4A=4+4^2+4^3+4^4+....+4^100

4A-A=4^100-1

=>3A=4^100-1 mà 4^100-1<4^100

=>3A<B  =>A<B/3(đpcm) 

Ta có: A = 1+4+4^2+4^3+...+4^99  
=> 4A = 4.(1+4+4^2+4^3+...+4^99)
=> 4A = 4+4^2+4^3+...+4^99+4^100 
=> 4A - A = (4+4^2+4^3+...+4^99+4^100) - (1+4+4^2+4^3+...+4^99) 
=> 3A = 4^100 - 1 
=> A = 4^100-1/3 < 4^100/3 mà B = 4^100 
=> A < 4^100/3 
Bài toán đã được chứng minh.

 

Ta có: \(A=1.3+2.4+3.5+4.6+...+99.101+100.102\)

\(A=1.\left(1+2\right)+2.\left(2+2\right)+3.\left(3+2\right)+4.\left(4+2\right)+....+99.\left(99+2\right)+100.\left(100+2\right)\)

\(A=\left(1^2+2^2+3^2+4^2+...+99^2+100^2\right)+\left(2+4+6+8+...+198+200\right)\)Đặt \(B=1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\)

\(\Rightarrow B=\left(1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\right)-2^2.\left(1^2+2^2+3^2+4^2+5^2+....+49^2+50^2\right)\)Tính dãy tổng quát \(C=1^2+2^2+3^2+4^2+5^2+...+n^2\)

\(C=1\left(0+1\right)+2\left(1+1\right)+3.\left(2+1\right)+4.\left(3+1\right)+5\left(4+1\right)+...+n\left[\left(n-1\right)+1\right]\)

\(C=\left[1.2+2.3+3.4+4.5+...+\left(n-1\right).n\right]+\left(1+2+3+4+5+....+n\right)\)

\(C=n.\left(n+1\right).\left[\left(n-1\right):3+1:2\right]=n.\left(n+1\right).\left(2n+1\right):6\)

Áp dụng vào B ta được:

\(B=100.101.201:6-4.50.51.101:6=166650\)

\(\Rightarrow A=166650+\left(200+2\right).100:2\)

\(\Rightarrow A=166650+10100=176750\)

Vậy A = 176750

Chúc bạn học tốt!!

Uk

A= 2^0 + 2^1 + 2^2 + ... + 2^2010

2A= 2^1 + 2^2 + ... + 2^2011

2A - A = (2^1 + 2^2 +... + 2^2011) - (2^0 + 2^1 + 2^2 +... 2^2010)

A = 2^2011 - 2^0

A = 2^2011 -1

=> A = B

OK Tick mình nghe