\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

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

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{99}\right)⋮3\)

Là \(\left(\dfrac{1}{2}\right)^2\) hay \(\dfrac{1}{2^2}\) vậy bạn

Những cái sau tương tự

\(\dfrac{1}{2^2}\)

s=[1+2]+[2+2 mũ 2]+...+[2 mũ 6+2 mũ 7]

s=1 nhân [1+2]+2 nhân [1+2]+...+2 mũ 6 nhân [1+2]

s=[1+2] nhân[1+2+...+2 mũ 6

s=3 nhân [1+2+...+2 mũ 6]

=> s chia hết cho 3

mong mn giúp mình gấp với ạ ^^

\(S=\left(2+2^2+2^3+2^4\right)+...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(=30\cdot\left(1+...+2^{96}\right)⋮10\)

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

\(=\left(2+2^2+2^3+2^4\right)+...+2^{95}\left(2+2^2+2^3+2^4\right)\)

\(=30\left(1+...+2^{95}\right)⋮10\)

a:Sửa đề: \(S=2+2^2+\cdots+2^{2024}\)

Ta có: \(S=2+2^2+\cdots+2^{2024}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{2023}+2^{2024}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{2023}\left(1+2\right)\)

\(=3\left(2+2^3+\cdots+2^{2023}\right)\) ⋮3

b: Ta có: \(S=2+2^2+\cdots+2^{2024}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{2021}+2^{2022}+2^{2023}+2^{2024}\right)\)

\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+\cdots+2^{2020}\left(2+2^2+2^3+2^4\right)\)

\(=30\left(1+2^4+\cdots+2^{2020}\right)=3\cdot10\cdot\left(1+2^4+\cdots+2^{2020}\right)\) ⋮10

=>S có chữ số tận cùng là 0

Lời giải:
$S=(2+2^2)+(2^3+2^4)+....+(2^{23}+2^{24})$

$=2(1+2)+2^3(1+2)+....+2^{23}(1+2)$

$=(1+2)(2+2^3+...+2^{23})$

$=3(2+2^3+...+2^{23})\vdots 3$

b.

$S=2+2^2+2^3+...+2^{23}+2^{24}$

$2S=2^2+2^3+2^4+....+2^{24}+2^{25}$

$\Rightarrow 2S-S=2^{25}-2$

$\Rightarrow S=2^{25}-2$

Ta có:

$2^{10}=1024=10k+4$

$\Rightarrow 2^{25}-2=2^5.2^{20}-2=32(10k+4)^2-2=32(100k^2+80k+16)-2$
$=10(320k^2+8k+51)\vdots 10$

$\Rightarrow S$ tận cùng là $0$

Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{118}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+\cdots+2^{118}\right)\) ⋮7

Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\cdots+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2\left(1+2+\cdots+2^4\right)+2^6\left(1+2+\cdots+2^4\right)+\cdots+2^{116}\left(1+2+\cdots+2^4\right)\)

\(=31\left(2+2^6+\ldots+2^{116}\right)\) ⋮31

Ta có: \(A=2+2^2+2^3+\cdots+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{119}+2^{120}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{119}\left(1+2\right)=3\left(2+2^3+\cdots+2^{119}\right)\) ⋮3

Ta có: A⋮3

A⋮7

mà ƯCLN(3;7)=1

nên A⋮3*7

=>A⋮21