2​A -A = 2 mũ 100-1 . vậy A bằng 2 mu 100-1. tích đúng giúp minh . k hiểu nhắn tin mk giảng

2 MŨ 100 - 1 = ?

A=1+2+2²+....+2¹⁰⁰

2A=2+2²+2³+...+2¹⁰¹

2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+....+2¹⁰⁰)

A=2¹⁰¹-1

Ban kia lam dung roi do 

k tui nha

thanks

bạn kia trả lời đúng đó

Đoàn Lê Thu Trang

Đặt x = 2k ( k thuộc N )

Số số hạng của dãy số đó là :

\(\frac{2k-2}{2}+1=k-1+1=k\) ( số hạng )

Do đó :

\(\frac{k\times\left(2k+2\right)}{2}=2450\)

\(k\times k+1=2450=49\times50\)

\(\Rightarrow k=49\)

Vậy x chỉ có thể bằng : 

49 x 2 = 98

b) 1+2-3-4+5+6-7-8+...-499-500+501+502

= (1+2-3-4)+(5+6-7-8)+...+(497+498-499-500)+501+502

= (-4)+(-4)+...+(-4)+501+502

= (-4.125)+501+502

= (-500)+501+502

= 503

Ax2=2+2²+2³+...+2⁸¹

Ax2-A=2⁸¹-2

A=2⁸⁰

Trả lời :

\(a,4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2002\)

\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)

\(=4.\left\{9.\left[33:11\right]-26\right\}+2002\)

\(=4.\left\{9.3-26\right\}+2002\)

\(=4.1+2002\)

\(=2006\)

Hok_tốt

#Thiên_Hy

\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2002\)

\(=4.\left\{9.\left[\left(25+8\right):11\right]-26\right\}+2002\)

\(=4.\left\{9.\left[33:11\right]-26\right\}+2002\)

\(=4.\left\{9.3-26\right\}+2002\)

\(=4.\left\{27-26\right\}+2002\)

\(=4.1+2002\)

\(=4+2002\)

\(=2006\)

~ Hok tốt ~

4.{3^2[(5^2+2^3):11]-26}+2002

=4.{9[(25+8):11]-26}+2002

=4.{9.33:11]-26}+2002

=4.{[297:11]-26}+2002

=4.{ 27 -16}+2002

=4.1+2002

=2+2002

=2008

a) = 25 x 32 + 25 x 47 - 32 x 25 + 32 x 47

   = 25 x 47 - 32 x 47

   = -329

b) = 2 - 12 - 6

    = -16

c) = 2 x (-7) - 10 + 1

    = -14 - 10 + 1

    = -23

d) = -6 - (-1) + (-2)

    = -7

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2018}{2^{2018}}\)

\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2018}{2^{2017}}\)

\(\Rightarrow2A-A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2018}{2^{2017}}-\frac{1}{2}-\frac{2}{2^2}-\frac{3}{2^3}-...-\frac{2018}{2^{2018}}\)

\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\frac{2018}{2^{2018}}\)

Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)

\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\)

\(\Rightarrow2B-B=2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-...-\frac{1}{2^{2017}}\)

\(\Rightarrow2B-B=2-\frac{1}{2^{2017}}\)

\(\Rightarrow A=2-\frac{2}{2^{2017}}-\frac{2018}{2^{2018}}\)

Thanks

Ok hok tốt

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2018}{2^{2017}}\)

\(\Rightarrow2A-A=\left(1+...+\frac{2018}{2^{2017}}\right)-\left(\frac{1}{2}+...+\frac{2018}{2^{2018}}\right)\)

\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}-\frac{2018}{2^{2018}}\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)

Tính 2B rùi trừ B  ( tương tự nhé )

\(\Rightarrow B=1-\frac{1}{2^{2017}}\)

\(\Rightarrow A=1+1-\frac{1}{2^{2017}}+\frac{2018}{2^{2018}}\)

\(\Rightarrow A=2-\frac{1}{2^{2017}}+\frac{2018}{2^{2018}}\)

\(\Rightarrow A=\frac{2^{2019}-2+2018}{2^{2018}}\)

ta có A=\(2^{2016}+2^{2015}+...+2^2+2^1+2^0\)

=\(\left(2-1\right)\left(2^{2016}+..+2^2+2^1+2^0\right)\)

=\(2^{2017}-1\)

áp dụng hằng đẳng thức 

\(A=1+2+2^2+2^3+...+2^{2006}\)

\(2A=2+2^2+2^3+....+2^{2007}\)

\(2A-A=2+2^2+2^3+..+2^{2007}-1-2-2^2-..-2^{2006}\)

\(A=2^{2007}-1\)

A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰⁰⁶ (1)

Nhân cả 2 vế của đẳng thức trên với 2 , ta được :

2A = 2¹ + 2² + 2³ + 2⁴ + .... + 2²⁰⁰⁷

  A = 2⁰ + 2¹ + 2² + 2³ + ... +  2²⁰⁰⁶

=> A = 2²⁰⁰⁶ - 2⁰

=> A = 2²⁰⁰⁶ - 1