Question

Chứng tỏ rằng

2 + 2² + 2³ + 2⁴ +.....+ 2⁹⁹ + 2¹⁰⁰ chia hết cho 3

Answer 1

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

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

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

\(=2.3+2^3.3+...+2^{99}.3\)

\(=3\left(2+2^3+...+2^{99}\right)\)chia hết cho 3 (Đpcm)

Answer 2

Đặt A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

Ta có:
A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

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

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

A = 2.3 + 2³.3 + ... + 2⁹⁹.3

A = (2 + 2³ + ... + 2⁹⁹) . 3

Vì (2 + 2³ + ... + 2⁹⁹) . 3 chia hết cho 3 nên 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰ chia hết cho 3 (đpcm)

Answer 3

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

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

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

\(\)\(=30+2^4\cdot30+...+2^{92}\cdot30\)

\(=30\cdot\left(2^4+2^8+...+2^{96}\right)\)

Vì 30 chia hết cho 3 \(\Rightarrow\)

\(2+2^2+2^3+2^4+...+2^{99}+2^{100}\)chia hết cho 3