Ta có

  2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 +...+ 2 98 + 2 99 + 2 100

= 2 1 + ( 2 2 + 2 3 + 2 4 ) + ( 2 5 + 2 6 + 2 7 ) +...+ ( 2 98 + 2 99 + 2 100 )

= 2 + 2 2 1 + 2 + 2 2 + 2 5 1 + 2 + 2 2 + . . . + 2 98 1 + 2 + 2 2

= 2 + 2 2 . 7 + 2 5 . 7 + . . . + 2 98 . 7 = 2 + 7 2 2 + 2 5 + . . . + 2 98

Mà  7 . 2 2 + 2 5 + . . . + 2 98 ⋮ 7  

Nên  2 + 7 2 2 + 2 5 + . . . + 2 98 : 7   d ư   2

Ta có A=2⁰+2¹+2²+2³+...2¹⁰⁰

2A=2¹+2²+...+2¹⁰¹

2A-A=(2¹+2²+...+2¹⁰⁰)-(2⁰+2¹+...+2¹⁰⁰)

A=2¹⁰¹-1

Mà 2¹⁰¹-1=(........02)-1=........01 chia 100 dư 1

Chúc bạn học tốt.

Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$

$=2.3+2^3.3+...+2^{99}.3$

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

Ta có đpcm.

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

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

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6\left(1+2^2+...+2^{98}\right)\)chia hết cho \(6\).

A=2+22+23+...+299+2100A=2+22+23+...+299+2100

⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101

⇒A=2101−2⇒A=2101−2

B=3+32+33+...+399+3100B=3+32+33+...+399+3100

⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101

⇒2B=3101−3⇒2B=3101−3

⇒B=3101−32

a: \(A=1+2+2^2+...+2^{41}\)

=>\(2A=2+2^2+2^3+...+2^{42}\)

=>\(2A-A=2^{42}-1\)

=>\(A=2^{42}-1\)

b: \(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{40}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{40}\right)⋮3\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{39}\left(1+2+2^2\right)\)

\(=7\left(1+2^3+...+2^{39}\right)⋮7\)