chua hoc den moi lop 7

+) Xét \(n=3k.\)

\(\Rightarrow2^n-1=2^{3k}-1=\left(2^3\right)^k-1=8^k-1=\left(8-1\right)\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]=7\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]⋮7\left(tm\right).\)

+) Xét \(n=3k+1.\)

\(\Rightarrow2^n-1=2^{3k+1}-1=2.8^k-1=2\left(8k-1\right)+1=2.\left(8-1\right)\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]+1=2.7\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]+1⋮̸7\left(loại\right).\)

+) Xét \(n=3k+2.\)

\(\Rightarrow2^n-1=2^{3k+2}-1=4.8^k-1=4\left(8^k-1\right)+3=4\left(8-1\right)\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]+3=4.7\left[\left(8^{k-1}\right)+\left(8^{k-2}\right)+...+1\right]+3⋮̸7\left(loại\right).\)

Vậy \(2^n-1⋮7\Leftrightarrow n=3k\) (k \(\in\) N^*).