A=\((1+2)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)

A=\(3.1+2^2\left(1+2\right)+...+2^{19}\left(1+2\right)\)

A=\(3.1+3.2^2+...+3.2^{19}\)

A=\(3\left(1+2^2+...+2^{19}\right)\)\(⋮3\)

Vậy A\(⋮3\)

A=

A=

A=

A=

NÊN  A

???

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/

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

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

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

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

A = 2 + 22 + 23 + 24 + ... + 219 + 220

A = (2 + 22) + (23 + 24) +... + (219 + 220)

A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)

A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3

do đó A chia hết cho 3

Sửa đề: \(A=2+2^2+2^3+2^4+...+2^{19}+2^{20}\)

=>\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{19}\right)⋮3\)

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)

\(A=2^{21}-2\)

___________

\(B=5+5^2+...+5^{50}\)

\(5B=5^2+5^3+...+5^{51}\)

\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)

\(4B=5^{51}-5\)

\(B=\dfrac{5^{51}-5}{4}\)

___________

\(C=1+3+3^2+...+3^{100}\)

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

\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)

\(2C=3^{101}-1\)

\(C=\dfrac{3^{101}-1}{2}\)

2A= 2(2+2²+2³+...+2¹⁹+2²⁰)

2A= 2²+2³+2⁴+...+2²⁰+2²¹

2A-A=(2²+2³+2⁴+...+2²⁰+2²¹)-(2+2²+2³+...+2¹⁹+2²⁰)

A=2²¹-2

Vậy A=2²¹-2

Làm tương tự nhee