a) \(S=1+2+2^2+...+2^{100}\)

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

\(2S-S=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)

\(S=2^{101}-1\)

b) \(X=2^{2012}-2^{2011}-...-2-1\)

\(X=2^{2012}-\left(1+2+...+2^{2011}\right)\)

Đặt \(X=2^{2012}-Y\)

Ta có :

\(Y=1+2+...+2^{2011}\)

\(2Y=2+2^2+...+2^{2012}\)

\(2Y-Y=\left(2+2^2+...+2^{2012}\right)-\left(1+2+...+2^{2011}\right)\)

\(Y=2^{2012}-1\)

\(\Rightarrow X=2^{2012}-2^{2012}+1\)

\(\Rightarrow X=1\)

\(\Rightarrow2010X=2010\)

Số các số hạng là:

(2100 - 1) : 1 + 1 = 2100 số

Tổng của dãy A là:

(2100 + 1) x 2100 : 2 = 2206050

Đáp số : 2206050

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3

2^2022 mà bạn chuche ơi

1) 2300 : 50 = ( 2300 . 2 ) : ( 50 . 2 ) = 4600 : 100 = 46

2) 112 : 2 = ( 112 . 5 ) : ( 2 . 5 ) = 560 : 10 = 56

3) 2300 : 4 = ( 2300 . 25 ) : ( 4 . 25 ) = 57500 : 100 = 575

4) 3500 : 25 = ( 3500 . 4 ) : ( 4 . 25 ) = 14000 : 100 = 140

5) 1200 : 75 = ( 1200 . 4 ) : ( 4 . 75 ) = 4800 : 300 = 6

6) 2100 : 75 = ( 2100 . 4 ) : ( 4 . 75 ) = 8400 : 300 = 28

7) 320 : 8 = ( 320 . 5 ) : ( 8 . 5 ) = 1600 : 40 = 40

1) 2300 : 50 = (2300.2) : (50.2) = 4600 : 100 = 46

2) 112: 2 = ( 112.5) : (2.5) = 560 : 10 = 56

3) 2300 : 4 = ( 2300.25) : (4.25) = 57500 : 100 = 575

4)3500 : 25 = (3500.4) : (25.4) = 14000 : 100 = 140

5)1200 : 75 = ( 1200 : 3) : (75:3) = 400 : 25 = (400.4) : (25.4) = 1600 : 100 = 16

...

1) 2300 : 50 = ( 2300 . 2 ) : ( 50 . 2 ) = 4600 : 100 = 46

2) 112 : 2 = ( 112 . 5 ) : ( 2 . 5 ) = 560 : 10 = 56

3) 2300 : 4 = ( 2300 . 25 ) : ( 4 . 25 ) = 57500 : 100 = 575

4) 3500 : 25 = ( 3500 . 4 ) : ( 4 . 25 ) = 14000 : 100 = 140

5) 1200 : 75 = ( 1200 . 4 ) : ( 4 . 75 ) = 4800 : 300 = 6

6) 2100 : 75 = ( 2100 . 4 ) : ( 4 . 75 ) = 8400 : 300 = 28

7) 320 : 8 = ( 320 . 5 ) : ( 8 . 5 ) = 1600 : 40 = 40

Cho A = 2^2 + 2^3  + 2^4 + ...+ 2^98 + 2^99 và B = 2^100 - 5. So sánh A và B

Bài giải

ta có: A = 2^2+2^3+2^4+...+2^98+2^99

=> 2A = 2^3+2^4+2^5+...+2^99+2^100

=> 2A-A = 2^100 - 2^2

A = 2^100 - 4 > B = 2^100 - 5

=> A >B

Ta có : A = 2² + 2³ + ...... + 2⁹⁹

=> 2A = 2³ + 2⁴ + ...... + 2¹⁰⁰

=> 2A  - A = 2¹⁰⁰ - 2² 

=> A = 2¹⁰⁰ - 4 > 2¹⁰⁰ - 5

Vậy A > B 

Ta có: A = 2² + 2³ + 2⁴ +...+ 2⁸⁹ + 2⁹⁹  
         2A = 2³ + 2⁴ + 2⁵ +...+ 2⁹⁰ + 2¹⁰⁰ 
     2A-A = (2³ + 2⁴ + 2⁵ +...+ 2⁹⁰ + 2¹⁰⁰ ) - (2² + 2³ + 2⁴ +...+ 2⁸⁹ + 2⁹⁹   )
           A = 2³ + 2⁴ + 2⁵ +...+ 2⁹⁰ + 2¹⁰⁰ - 2² - 2³ - 2⁴ -...- 2⁸⁹  -2⁹⁹  
           A = 2⁹⁰ + 2¹⁰⁰ - 2² - 2⁸⁹ - 2⁹⁹ 
 

Suy ra A > B
Chúc bạn học tốt

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

     \(=1-\frac{1}{100}=\frac{99}{100}\)

nhìu dữ, nhưg liệu có ****?

1/

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

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

\(\Rightarrow S=A-2B\)

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

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

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe