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Câu 5 nhé

A = 1/2 + 1/2² + 1/2³ + 1/2⁴+...+ 1/2²⁰²¹ + 1/2²⁰²²

và B = 1/3+1/4+1/5+17/60

Hỏi :

a) Rút gọn A

  b)So sánh A và B

a) \(A=2A-A\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1-\dfrac{1}{2^{2022}}\)

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )

A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60

Ta có: $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}}$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}$

$\Rightarrow 2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}})$

$\Rightarrow A=1-\frac{1}{2^{2022}}<1$

$\Rightarrow A<1\left(1\right)$

Lại có: $B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{17}{60}$

$\Rightarrow B=\frac{16}{15}$

$\Rightarrow B=1+\frac{1}{15}>1$

$\Rightarrow B>1\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)\Rightarrow A<B$

Vậy $A<B$

2A=2*(1+2+2²+...+2²⁰²⁰)=2+2²+...+2²⁰²¹

^2A-A=(1+2+2²+...+2²⁰²¹)-(1+2+2²+...+2²⁰²⁰)

A=2²⁰²¹-1<2021

Giải:

A=1+2+2²+2³+...+2²⁰²⁰

2A=2+2²+2³+2⁴+...+2²⁰²¹

2A-A=(2+2²+2³+2⁴+...+2²⁰²¹)-(1+2+2²+2³+...+2²⁰²⁰)

A=2²⁰²¹-1

⇒A<2²⁰²¹

Chúc bạn học tốt!

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

Bài 1

a, cm : A = 16⁵ + 2¹⁵ ⋮ 3

    A = 16⁵ + 2¹⁵

   A = (2⁴)⁵ +  2¹⁵

  A  = 2²⁰ + 2¹⁵

 A  =  2¹⁵.(2⁵ + 1)

 A = 2¹⁵. 33 ⋮ 3 (đpcm)

b,cm : B = 8⁸ + 2²⁰ ⋮ 17

    B = (2³)⁸ + 2²⁰ 

    B =  2¹⁶ + 2²⁰

    B = 2¹⁶.(1 + 2⁴)

    B = 2¹⁶. 17 ⋮ 17 (đpcm)

c, cm: C = 1 - 2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶ -...-2²⁰²¹ + 2²⁰²² : 6 dư 1

C=1+(-2+2²-2³+2⁴- 2⁵+2⁶)+...+(-2²⁰¹⁷+2²⁰¹⁸-2²⁰¹⁹+2²⁰²⁰-2²⁰²¹+2²⁰²²)

C = 1 + 42 +...+ 2²⁰¹⁶.(-2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶)

C = 1 + 42+...+ 2²⁰¹⁶.42

C = 1 + 42.(2⁰+...+2²⁰¹⁶)

42 ⋮ 6 ⇒ C = 1 + 42.(2⁰+...+2²⁰¹⁶) : 6 dư 1 đpcm

a, \(\overline{aaa}\) \(⋮\) 37

    \(\overline{aaa}\) = a x 111 = a x 3 x 37 ⋮ 37 (đpcm)

b, (\(\overline{ab}\) + \(\overline{ba}\)) ⋮ 11

  \(\overline{ab}\) + \(\overline{ba}\) = \(\overline{a0}\) + b + \(\overline{b0}\) + a = \(\overline{aa}\) + \(\overline{bb}\) = a x 11 + b x 11 = 11 x (a+b)⋮11

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