A number theory problem by Wildan Bagus Wicaksono

2^n + 1 .cong. 0 (mod 3) if n is odd. So we require the sum of all odd integers , 0 < n < 101. This is (50/2) (1 + 99) = 25 100 = 2500.

$\large { (3-1) }^{ n }+1=\left\{ \sum _{ i=0 }^{ n }{ \left( \begin{matrix} n \\ i \end{matrix} \right) } { 3 }^{ n-i }\cdot { \left( -1 \right)^i } \right\} +1\\ { (3-1) }^{ n }+1=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { 3 }^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) \cdot { 3 }^{ n-1 }\cdot { \left( -1 \right) }^{ 1 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) 3{ \left( -1 \right) }^{ n-1 }+\left( \begin{matrix} n \\ n \end{matrix} \right) { \left( -1 \right) }^{ n }+1\\ { (3-1) }^{ n }+1=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { 3 }^{ n }-\left( \begin{matrix} n \\ 1 \end{matrix} \right) \cdot { 3 }^{ n-1 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) 3{ \left( -1 \right) }^{ n-1 }+\left( \begin{matrix} n \\ n \end{matrix} \right) { \left( -1 \right) }^{ n }+1$

The two decisive expressions ${ 2 }^{ n }+1$ divisible by $3$ are ${ (-1) }^{ n }+1$ must be $0$ , meaning $n$ must be an odd number because $(-1) + 1 = 0$ . Thus, ${ 2 }^{ n }+1$ is divided by $3$ . Thus, the sum of all $n$ values with $n<100$ ,

$1+3+5+...+99={ \left( \frac { 1+99 }{ 2 } \right) }^{ 2 }\\ 1+3+5+...+99={ 50 }^{ 2 }\\ 1+3+5+...+99=2500$