No tengo ni idea
Demuestre que para cualquier $n$ número natural esta suma $$ \sum\limits_ {k=0}^{n} \binom {2n+1}{2k+1}2^{3k}$$ no es divisible por $5$ .
$ \begin {array}{l} \left ( {1 + x} \right )^{2n + 1} - \left ( {1 - x} \right )^{2n + 1} = \sum\limits_ {k = 0}^{2n + 1} { \left ( { \begin {array}{*{20}c} {2n + 1} \\ k \\ \end {array}} \right )} x^k - \sum\limits_ {k = 0}^{2n + 1} { \left ( { \begin {array}{*{20}c} {2n + 1} \\ k \\ \end {array}} \right )} \left ( { - x} \right )^k = \sum\limits_ {k = 0}^n { \left ( { \begin {array}{*{20}c} {2n + 1} \\ {2k + 1} \\ \end {array}} \right )} x^{1 + k} \\ x = 2 \Rightarrow 3^{2n + 1} + 1 = \sum\limits_ {k = 0}^{2n + 1} { \left ( { \begin {array}{*{20}c} {2n + 1} \\ k \\ \end {array}} \right )} 2^k - \sum\limits_ {k = 0}^{2n + 1} { \left ( { \begin {array}{*{20}c} {2n + 1} \\ k \\ \end {array}} \right )} \left ( { - 2} \right )^k \\ 3^{2n + 1} + 1 = \sum\limits_ {k = 0}^n { \left ( { \begin {array}{*{20}c} {2n + 1} \\ {2k + 1} \\ \end {array}} \right )} 2^{k + 1} \\ \end {array}$
$ \begin {array}{l} \left\ { \begin {array}{l} 3^0 \equiv 1 \left [ 5 \right ] \\ 3^1 \equiv 3 \left [ 5 \right ] \\ 3^2 \equiv 4 \left [ 5 \right ] \\ 3^3 \equiv 2 \left [ 5 \right ] \\ 3^4 \equiv 1 \left [ 5 \right ] \\ \end {array} \right. \Rightarrow \forall m \in\mathbb N:\left\ { \begin {array}{l} 3^{4m} \equiv 1 \left [ 5 \right ] \\ 3^{4m + 1} \equiv 3 \left [ 5 \right ] \\ 3^{4m + 2} \equiv 4 \left [ 5 \right ] \\ 3^{4m + 3} \equiv 2 \left [ 5 \right ] \\ \end {array} \right. \\ 3^{2 \left ( {2m} \right ) + 1} + 1 \equiv 2 \left [ 5 \right ]and3^{2 \left ( {2m + 1} \right ) + 1} + 1 \equiv 3 \left [ 5 \right ] \\ \end {array} $
gracias de antemano