정수 부분의 합

$\displaystyle \sum\limits _{k=1} ^{10} f ( 2 ^{k} ) = \sum\limits _{k=1} ^{10} 2 ^{k}$$\displaystyle = \frac{2 ( 2 ^{10} - 1 )}{2 - 1}$$= 2046$ $2 ^{n} \leq k < 2 ^{n+1}$일 때, $n \leq \log _{2} k < n + 1$이므로 $f \left( \log _{2} k \right) = n$이다. 이때, $f \left( \log _{2} k \right) = n$인 자연수 $k$의 개수는 $2 ^{n+1} - 2 ^{n} = 2 ^{n}$이다. $2 = 2 ^{1}$, $1024 = 2 ^{10}$이므로 $\displaystyle \sum\limits _{k=2} ^{1024} f \left( \log _{2} k \right)$$\displaystyle = \sum\limits _{n=1} ^{9} n \cdot 2 ^{n} + 10$ $S$$\displaystyle = \sum\limits _{n=1} ^{9} n \cdot 2 ^{n}$$= 1 \cdot 2 + 2 \cdot 2 ^{2} + 3 \cdot 2 ^{3} + \cdots + 9 \cdot 2 ^{9}$ ⋯⋯ ㉠ 이라 하면 $2 S$$\displaystyle = \sum\limits _{n=1} ^{9} n \cdot 2 ^{n+1}$$= 1 \cdot 2 ^{2} + 2 \cdot 2 ^{3} + 3 \cdot 2 ^{4} + \cdots + 9 \cdot 2 ^{10}$ ⋯⋯ ㉡ ㉠$-$㉡에서 $\displaystyle - S = \sum\limits _{n=1} ^{9} 2 ^{n} - 9 \cdot 2 ^{10}$

∴ $S$$\displaystyle = 9 \cdot 2 ^{10} - \sum\limits _{n=1} ^{9} 2 ^{n}$ $\displaystyle = 9 \cdot 2 ^{10} - \frac{2 ( 2 ^{9} - 1 )}{2 - 1}$ $= 8 \cdot 2 ^{10} + 2$ $= 8194$ ∴ $\displaystyle \sum\limits _{k=2} ^{1024} f \left( \log _{2} k \right)$$= 8194 + 10 = 8204$ ∴ $\displaystyle \sum\limits _{k=1} ^{10} f ( 2 ^{k} ) + \sum\limits _{k=2} ^{1024} f \left( \log _{2} k \right)$$= 2046 + 8204$$= 10250$