Nacional 2014 N3 P4

[LuchoLP]

Consideremos las sumas de $50$ sumandos

$S=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{99\cdot 100}$,

$T=\frac{1}{51\cdot 100}+\frac{1}{52\cdot 99}+\ldots +\frac{1}{99\cdot 52}+\frac{1}{100\cdot 51}$.

Expresar el cociente $\frac{S}{T}$ como una fracción irreducible.

[JPablo]

Spoiler: mostrar

A través de una calculadora simple podemos obtener un resultado aproximado. Dependiendo de si la calculadora posee o no la función sumatoria, el resultado será más aproximado.

Sea cual sea el método utilizado, podemos especular que el resultado será igual a $\frac{151}{2}$, por lo que lo que debemos hacer es demostrar este resultado.

Notemos que

${S=\sum \limits _{k=1}^{50}\frac{1}{2k\left (2k-1\right )}=\sum \limits _{k=1}^{50}\frac{\left (2k\right )-\left (2k-1\right )}{2k\left (2k-1\right )}=\sum \limits _{k=1}^{50}\left (\frac{1}{2k-1}-\frac{1}{2k}\right )=\sum \limits _{k=1}^{50}\frac{1}{2k-1}-\sum_{k=1}^{50}\frac{1}{2k}}$

$T=\sum \limits _{k=1}^{50}\frac{1}{\left (k+50\right )\left (101-k\right )}$

Si se cumpliera nuestro resultado especulado, tendríamos

$S=\frac{151}{2}T=\frac{151}{2}\sum \limits _{k=1}^{50}\frac{1}{\left (k+50\right )\left (101-k\right )}=\frac{1}{2}\sum \limits _{k=1}^{50}\frac{151}{\left (k+50\right )\left (101-k\right )}$

Pero notemos que la sucesión $\frac{151}{\left (k+50\right )\left (101-k\right )}$ es simétrica, entonces

$\sum \limits _{k=1}^{50}\frac{151}{\left (k+50\right )\left (101-k\right )}=2\sum \limits _{k=1}^{25}\frac{151}{\left (k+50\right )\left (101-k\right )}$

Por lo que queda entonces que

$S=\sum \limits _{k=1}^{25}\frac{151}{\left (k+50\right )\left (101-k\right )}$

Reemplazamos $151=\left (k+50\right )+\left (101-k\right )$ y nos queda

${S=\sum \limits _{k=1}^{25}\frac{\left (k+50\right )+\left (101-k\right )}{\left (k+50\right )\left (101-k\right )}=\sum \limits _{k=1}^{25}\left (\frac{1}{101-k}+\frac{1}{k+50}\right )=\sum \limits _{k=1}^{25}\frac{1}{101-k}+\sum \limits _{k=1}^{25}\frac{1}{k+50}}$

Notemos que

$\sum \limits _{k=1}^{25}\frac{1}{101-k}=\frac{1}{100}+\frac{1}{99}+\cdots +\frac{1}{76}$

$\sum \limits _{k=1}^{25}\frac{1}{k+50}=\frac{1}{51}+\frac{1}{52}+\cdots +\frac{1}{75}$

Por lo tanto

$S=\sum \limits _{k=1}^{25}\frac{1}{101-k}+\sum \limits _{k=1}^{25}\frac{1}{k+50}=\sum \limits _{k=51}^{100}\frac{1}{k}=\sum \limits _{k=1}^{50}\frac{1}{k+50}$

Es decir que debemos demostrar que

$\sum \limits _{k=1}^{50}\frac{1}{2k-1}-\sum \limits _{k=1}^{50}\frac{1}{2k}=\sum \limits _{k=1}^{50}\frac{1}{k+50}\Longleftrightarrow \sum \limits _{k=1}^{50}\frac{1}{2k-1}=\sum \limits _{k=1}^{50}\frac{1}{k+50}+\frac{1}{2}\sum \limits _{k=1}^{50}\frac{1}{k}$

Es decir, demostrar que

$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{99}=\left (\frac{1}{51}+\frac{1}{52}+\cdots +\frac{1}{100}\right )+\frac{1}{2}\left (\frac{1}{1}+\frac{1}{2}+\cdots +\frac{1}{50}\right )$

De esta relación, podemos cancelar los términos iguales a ambos lados de la igualdad, y queda

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{49}=\left (\frac{1}{52}+\frac{1}{54}+\frac{1}{56}+\cdots +\frac{1}{100}\right )+\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{100}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{49}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{50}\right )+2\left (\frac{1}{52}+\frac{1}{54}+\frac{1}{56}+\cdots +\frac{1}{100}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{49}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{50}\right )+2\sum \limits _{k=26}^{50}\frac{1}{2k}$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{49}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{50}\right )+\sum \limits _{k=26}^{50}\frac{1}{k}$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{49}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{50}\right )+\left (\frac{1}{27}+\frac{1}{28}+\cdots +\frac{1}{50}\right )$

Volvemos a cancelar términos iguales, y repetimos el procedimiento:

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{25}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{50}\right )+\left (\frac{1}{28}+\frac{1}{30}+\frac{1}{32}+\cdots +\frac{1}{50}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{25}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{26}\right )+2\left (\frac{1}{28}+\frac{1}{30}+\frac{1}{32}+\cdots +\frac{1}{50}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{25}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{26}\right )+\left (\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\cdots +\frac{1}{25}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{13}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{26}\right )+\left (\frac{1}{14}+\frac{1}{16}+\frac{1}{18}+\cdots +\frac{1}{24}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{13}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{12}\right )+2\left (\frac{1}{14}+\frac{1}{16}+\frac{1}{18}+\cdots +\frac{1}{24}\right )$

$\frac{1}{1}+\frac{1}{3}+\cdots +\frac{1}{13}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{12}\right )+\left (\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots +\frac{1}{12}\right )$

$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots +\frac{1}{12}\right )+\left (\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right )$

$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right )+2\left (\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right )$

$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right )+\left (\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right )$

$\frac{1}{1}+\frac{1}{3}=\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{6} \right )+\left (\frac{1}{4}+\frac{1}{6}\right )$

Y es fácil verificar que esta igualdad es cierta, por lo tanto es cierto que

$\frac{S}{T}=\frac{151}{2}$

[mszew]

Spoiler: mostrar

Sea $S_1=\frac{1}{1\cdot 2}$
$S_2=S_1+\frac{1}{3\cdot 4}$
$\vdots$
$S_{i+1}=S_i+\frac{1}{(2i+1)\cdot 2(i+1)}$
Similarmente sea $T_1=\frac{1}{2\cdot 2}$
$T_2=\frac{1}{3.4}+\frac{1}{4\cdot 3}=\frac{1}{6}$
$T_3=\frac{1}{4.6}+\frac{1}{5\cdot 5}+\frac{1}{6\cdot 4}$
$\vdots$
$T_i=\frac{1}{(i+1)\cdot 2i}+\frac{1}{(i+2)\cdot (2i-1)}+\cdots+\frac{1}{2i\cdot (i+1)}$

Sea $R_i=\frac{S_i}{T_i}$
Casos pequeños:
$R_1=\frac{\frac{1}{2}}{\frac{1}{4}}=2=\frac{4}{2}$, $R_2=\frac{\frac{7}{12}}{\frac{1}{6}}=\frac{7}{2}$, $R_3=\frac{\frac{37}{60}}{\frac{37}{300}}=5=\frac{10}{2}$

Sera cierto que $R_i=\frac{3i+1}{2}$ o $2 S_i=(3i+1)T_i$?
Probemos inducción:
Se cumple para $i=1$:
Asumimos que: $2S_i=(3i+1)T_i$

Veamos cuanto da: $2S_{i+1}=2\left (S_i+\frac{1}{(2i+1)\cdot 2(i+1)}\right )=(3i+1)T_i+\frac{1}{(2i+1)\cdot (i+1)}$

Pero notemos que:
$(3i+1)T_i=\frac{(i+1)+2i}{(i+1)\cdot 2i}+\frac{(i+2)+(2i-1)}{(i+2)\cdot (2i-1)}+\cdots +\frac{(2i)+(i+1)}{2i\cdot (i+1)}$
$(3i+1)T_i=\frac{1}{2i}+\frac{1}{i+1}+\frac{1}{2i-1}+\frac{1}{i+2}+\cdots +\frac{1}{i+1}+\frac{1}{2i}$
$(3i+1)T_i=2\left (\frac{1}{i+1}+\frac{1}{i+2}+\cdots +\frac{1}{2i}\right )$
Similarmente:
$(3(i+1)+1)T_{i+1}=2\left (\frac{1}{i+2}+\frac{1}{i+3}+\cdots +\frac{1}{2i+2}\right )=$
$=2\left (\frac{1}{i+1}+\frac{1}{i+2}+\cdots +\frac{1}{2i}\right )+2\left (\frac{1}{2i+1}+\frac{1}{2i+2}-\frac{1}{i+1}\right )=$
$=(3i+1)T_i+\frac{1}{(2i+1)\cdot (i+1)}$
Entonces QED

Para el caso particular de $R_{50}=\frac{151}{2}$

[Nowhereman]

Una sin inducción y sin calculadora

Spoiler: mostrar

Expresando $S$ y $T$ como sumatorias y trabajando las expresiones tenemos que

$S=\sum \limits _{k=1}^{50}\frac{1}{(2k-1)2k}=\sum \limits _{k=1}^{50}\frac{1}{2k-1}-\frac{1}{2k}=\sum \limits _{k=1}^{50}\frac{1}{2k-1}-\frac{1}{2}\sum \limits _{k=1}^{50}\frac{1}{k}$

Notemos que $\sum \limits _{k=1}^{50}\frac{1}{2k-1}=\sum \limits _{k=1}^{100}\frac{1}{k}-\sum \limits _{k=1}^{50}\frac{1}{2k}=\sum \limits _{k=1}^{100}\frac{1}{k}-\frac{1}{2}\sum \limits _{k=1}^{50}\frac{1}{k}$

Reemplazando tenemos que

$S=\sum \limits _{k=1}^{100}\frac{1}{k}-\frac{1}{2}\sum \limits _{k=1}^{50}\frac{1}{k}-\frac{1}{2}\sum \limits _{k=1}^{50}\frac{1}{k}=\sum \limits _{k=1}^{100}\frac{1}{k}-\sum \limits _{k=1}^{50}\frac{1}{k}$

Ahora viendo $T$

$T=\sum \limits _{k=1}^{50}\frac{1}{(50+k)(101-k)}=\sum \limits _{k=1}^{50}\left (\frac{1}{50+k}+\frac{1}{101-k}\right )\frac{1}{151}$

notemos que $\sum \limits _{k=1}^{50}\frac{1}{50+k}=\sum \limits _{k=1}^{50}\frac{1}{101-k}=\sum \limits _{k=51}^{100}\frac{1}{k}=\sum \limits _{k=1}^{100}\frac{1}{k}-\sum \limits _{k=1}^{50}\frac{1}{k}=S$

Reemplazando en $T$ tenemos que

$T=\frac{2}{151}S$ entonces $\frac{S}{T}=\frac{151}{2}$.

[biank]

Spoiler: mostrar

$S = \sum_{x = 1}^{50} \frac{1}{(2x - 1)2x} = \sum_{x=51}^{100} \frac{1}{x}$

Spoiler: mostrar

$\frac{1}{(2x - 1)2x} = \frac{1}{2x - 1} - \frac{1}{2x}$

Spoiler: mostrar

Por fracciones simples
$\frac{1}{(2x - 1)2x} = \frac{A}{2x - 1} + \frac{B}{2x}$
$1 = A2x + B(2x - 1)$

Reemplazando con $x = 0$
$1 = -B \Rightarrow B = -1$

Reemplazando con $x = \frac{1}{2}$
$1 = A$

Entonces $S = \sum_{x = 1}^{50} \frac{1}{2x - 1} - \sum_{x = 1}^{50} \frac{1}{2x}$

$\sum_{x = 1}^{50} \frac{1}{2x - 1} = \sum_{x = 1}^{100} \frac{1}{x} - \sum_{x = 1}^{50} \frac{1}{2x}$

$\Rightarrow S = \sum_{x = 1}^{100} \frac{1}{x} - 2 \sum_{x=1}^{50} \frac{1}{2x} = \sum_{x = 1}^{100} \frac{1}{x} - \sum_{x=1}^{50} \frac{1}{x} = \sum_{x=51}^{100} \frac{1}{x}$

$T = \sum_{x = 1}^{50} \frac{1}{(x + 50)(101 - x)} = \frac{2}{151} \sum_{x = 51}^{100} \frac{1}{x}$

Spoiler: mostrar

$\frac{1}{(x + 50)(101 - x)} = \frac{1}{151}\left(\frac{1}{x + 50} + \frac{1}{101 - x}\right)$

Spoiler: mostrar

Por fracciones simples
$\frac{1}{(x + 50)(101 - x)} = \frac{A}{x + 50} + \frac{B}{101 - x}$
$1 = A(101 - x) + B(x + 50)$

Reemplazando con $x = 101$
$1 = 151B \Rightarrow B = \frac{1}{151}$

Reemplazando con $x = -50$
$1 = 151A \Rightarrow A = \frac{1}{151}$

Entonces $T = \frac{1}{151}\left(\sum_{x = 1}^{50} \frac{1}{x + 50} + \sum_{x = 1}^{50} \frac{1}{101 - x}\right) = \frac{1}{151}\left(\sum_{x = 51}^{100} \frac{1}{x} + \sum_{x = 51}^{100} \frac{1}{x}\right) = \frac{2}{151} \sum_{x = 51}^{100} \frac{1}{x}$

$\dfrac{S}{T} = \dfrac{\sum_{x = 51}^{100} \frac{1}{x}}{\frac{2}{151} \sum_{x = 51}^{100} \frac{1}{x}} = \dfrac{151}{2}$