Divisor Sum of Power of 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{>0}$ be a power of $2$.

Then:

$\map {\sigma_1} n = 2 n - 1$


Proof

Let $n = 2^k$.

Then:

\(\ds \map {\sigma_1} n\) \(=\) \(\ds \dfrac {2^{k + 1} - 1} {2 - 1}\) Divisor Sum of Power of Prime
\(\ds \) \(=\) \(\ds 2 \times 2^k - 1\)
\(\ds \) \(=\) \(\ds 2 n - 1\)

$\blacksquare$