Question

[friend forever II Lê Tiến Đạt]

S=\(2+2^2+2^3+...+2^{10}\)

Chứng minh rằng 

- S\(⋮\)3

- S \(⋮\)31

Answer 1

[cô gái bọ cạp]

+ ) \(S=2+2^2+2^3+...+2^{10}\)

\(S=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(S=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^9.\left(1+2\right)\)

\(S=2.3+2^3.3+...+2^9.3\)

\(S=3.\left(2+2^3+...+2^9\right)\)

Vì \(3⋮3\)nên \(3.\left(2+2^3+...+2^9\right)⋮3\)hay\(S⋮3\)

Answer 2

[Edogawa Conan]

Chứng minh: S \(⋮\)3

Ta có: \(S=2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+....+2^9\left(1+2\right)\)

\(=2\times3+2^3\times3+....+2^9\times3\)

\(=3\times\left(2+2^3+...+2^9\right)\)

Vậy S chia hết cho 3 (vì có chứa thừa số 3)

Chứng minh: \(S⋮31\)

Ta có: S = \(\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)

\(=2\times31+2^6\times31\)

\(=31\times\left(2+2^6\right)\)

Vậy S chia hết cho 31 (vì có chứa thừa số 31)