Ta có:

A = 2 + 2² + 2³ + ... + 2¹⁰⁰

A = (2 + 2²) + (2³ + 2⁴) + ... + (2⁹⁹ + 2¹⁰⁰)

A = 2(1 + 2) + 2³(1 + 2) + ... + 2⁹⁹(1 + 2)

A = 3 × (2 + 2³ + ... + 2⁹⁹)

Vì 2 + 2³ + ... + 2⁹⁹ là số chẵn nên A chia hết cho 6.

Mặt khác:

A = 2 + 2² + 2³ + ... + 2¹⁰⁰

A = 2(1 + 2 + 2² + ... + 2⁹⁹)

Vì \(2^4=16\) và \(2^4-1=15\), ta có thể xét theo modulo 30:

\(2^4\equiv16\pmod{30}\).

Tuy nhiên, A không chia hết cho 30 vì A có số dư 2 khi chia cho 5.

Vậy đề bài phần a có thể bị nhầm: A chia hết cho 6 nhưng không chia hết cho 30.

Ta có:

B = 5 + 5² + 5³ + ... + 5⁹⁶

B = 5(1 + 5 + 5² + ... + 5⁹⁵)

Vì 5 không chia hết cho 6 nên cần kiểm tra lại đề bài. Thực tế B không chia hết cho 6.

Ta có:

A = 2 + 2² + 2³ + ... + 2¹⁰⁰

A = 2(2¹⁰⁰ - 1)

A = 2¹⁰¹ - 2

Với B:

B = 5 + 5² + 5³ + ... + 5⁹⁶

B = 5(5⁹⁶ - 1) : (5 - 1)

B = \(\dfrac{5^{97}-5}{4}\)

Với C:

C = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2

C = (2¹⁰⁰ - 2⁹⁹) + (2⁹⁸ - 2⁹⁷) + ... + (2² - 2)

C = 2⁹⁹ + 2⁹⁷ + ... + 2

C = 2(1 + 2² + 2⁴ + ... + 2⁹⁸)

C = \(\dfrac{2(2^{100}-1)}{3}\)

ukm

it 's very hard

l can do it 

sorry!

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

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

\(A=\left(1+4\right).\left(4\right)+\left(1+4\right).\left(4^3\right)+...+\left(1+4\right).\left(4^{99}\right)\)

\(A=5.\left(4+4^3+4^5+...+4^{99}\right)\)

Vậy A chia hết cho 5

Các bạn nha!

A=2+2²+2³+....+2⁹⁹+2¹⁰⁰

A=(2+2²+2³+2⁴+2⁵)+(2⁶+2⁷+2⁸+2⁹+2¹⁰)+......+(2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

A=(2+2²+2³+2⁴+2⁵)+2⁵.(2+2²+2³+2⁴+2⁵)+....+2⁹⁵.(2+2²+2³+2⁴+2⁵)

A=62+2⁵.62+.....+2⁹⁵.62

A=62.(1+2⁵+.....+2⁹⁵)

A=31.2.(1+2⁵+...+2⁹⁵)\(⋮\)31

Vậy A\(⋮\)31

Chúc bn học tốt

A=2+2^2+2^3+...+2^100

  =(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+....+(2^96+2^97+2^98+2^99+2^100)

=62+2^5(2+2^2+2^3+2^4+2^5)+....+2^95(2+2^2+2^3+2^4+2^5)

=62+2^5.62+2^10.62+....+2^95.62

=62(1+2^5+2^10+...+2^95)

Vì 62 chia hết cho 31 => A chia hết cho 31

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Cho a là số tự nhiênchia 6 dư 2 và b là số tự nhiên chia 6 dư 3. Chứng minh axb chia hết cho 6

1) Chứng minh \(ab\) chia hết cho \(6\)

Vì $a$ chia cho $6$ dư $2$ nên $a=6k+2$ (k thuộc Z).

Vì $b$ chia cho $6$ dư $3$ nên $b=6m+3$ (k thuộc Z).

Ta có:

$ab=(6k+2)(6m+3)$

$=36km+18k+12m+6$

$=6(6km+3k+2m+1)$

Vậy $ab$ chia hết cho $6$.

Ta có: A= 2+2^2+2^3+2^4+...+2^100=

= (2+2^2)+(2^3+2^4)+...+(2^99+2^100)=6+2^2.(2+2^2)+...+2^98.(2+2^2)=

= 6+2^2.6+...+2^98.6=6.(1+2^2+...+2^98)

Vì 6\(⋮\)6 nên 6.(1+2^2+...+2^98)\(⋮\)6 hay A\(⋮\)6.

nhớ k cho mk nhá, điểm mk đang âm,cảm ơn nhiều!!!

bài 1 

a)Số tận cùng là 6 nha