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.
b) Chứng tỏ B chia hết cho 6, 31, 26, 126Ta 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.
c) Tính giá trị A, B, CTa 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 = \(\dfrac{2(2^{100}-1)}{3}\)
\(A=4+4^2+4^3+...+4^{100}\)
\(A=\left(4+\text{ }4^2\right)+\left(4^3+4^4\right)+...+\left(4^{99}+4^{100}\right)\)
\(A=\left(1+4\right).\left(4\right)+\left(1+4\right).\left(4^3\right)+...+\left(1+4\right).\left(4^{99}\right)\)
\(A=5.\left(4+4^3+4^5+...+4^{99}\right)\)
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
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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.
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\\ A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)=6\left(1+2^2+...+2^{98}\right)⋮6\)
A=2+22+23+24+....+2100A=2+22+23+24+....+2100
A=(2+22)+(23+24)+....+(299+2100)A=(2+22)+(23+24)+....+(299+2100)
A=1.(2+22)+22.(2+22)+....+298.
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A=2+22+23+24+....+2100A=2+22+23+24+....+2100
A=(2+22)+(23+24)+....+(299+2100)A=(2+22)+(23+24)+....+(299+2100)
A=1.(2+22)+22.(2+22)+....+298.
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