Question

2+2²+2³+2⁴+.....+2²⁹+2³⁰

Chứng tỏ nó chia hết cho 3 và 7

Help Me!!!

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Answer 1

\(2+2^2+2^3+2^4+...\)\(+2^{29}+2^{30}\)

=\(2+\left(2.2\right)+\left(2.2^2\right)+\left(2.2^3\right)+...+\left(2.2^{28}\right)+\left(2.2^{29}\right)\)

=2.(1+2+2²+2³+...+2²⁸+2²⁹)

=2.(1+1073741822)

=2.1073741823

=2147483646

Answer 2

Ta có:

\(2+2^2+2^3+2^4+...+2^{29}+2^{30}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{29}+2^{30}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{29}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{29}.3\)

\(\Rightarrow\left(2+2^2+2^3+...+2^{30}\right)⋮3\)

Ta lại có:

\(2+2^2+2^3+2^4+...+2^{29}+2^{30}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{28}+2^{29}+2^{30}\right)\)

\(=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{28}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{28}.7\)

\(\Rightarrow\left(2+2^2+2^3+2^4+...+2^{29}+2^{30}\right)⋮7\)