[size=150][b]Practica 2. Ejercicio 10. iv)[/b][/size]
Hola. Alguien sabe como resolver este ejercicio? 

[tex]\sum_{k=0}^{n}{k.\binom{n}{k}}  = n.2^{n-1}[/tex]

HI: [tex]\sum_{i=0}^{k}{i.\binom{k}{i}}  = k.2^{k-1}[/tex]

Quiero probar que  [tex]\sum_{i=0}^{k+1}{i.\binom{k+1}{i}}  = (k+1).2^k[/tex]

Podria empezar asi?:
[tex]\sum_{i=0}^{k+1}{i.\binom{k+1}{i}} =[/tex]

[tex]\sum_{i=0}^{k}{i.\binom{k+1}{i}} + (k+1)\binom{k+1}{k+1} =[/tex] 

[tex]\sum_{i=1}^{k}{i.\binom{k+1}{i}} + 0 + (k+1)=[/tex]

[tex]\sum_{i=1}^{k}{i.\left[\binom{k}{i}-\binom{k}{i-1}\right]} + (k+1)=[/tex]

[tex]\sum_{i=0}^{k}{i.\binom{k}{i}} -\sum_{i=1}^{k}{i.\binom{k}{i-1}} + (k+1)=[/tex] 

por hipotesis inductiva....

[tex]k.2^{k-1} -\sum_{i=1}^{k}{i.\binom{k}{i-1}} + (k+1)=[/tex]

Si hago cambio de indice j = i - 1 en la sumatoria

[tex]\sum_{i=1}^{k}{i.\binom{k}{i-1}} =[/tex]

[tex]\sum_{j=0}^{k-1}{(j+1).\binom{k}{j}} =[/tex] 

[tex]\sum_{j=0}^{k-1}{j.\binom{k}{j}} + \sum_{j=0}^{k-1}{\binom{k}{j}} =[/tex] 

[tex]\sum_{j=0}^{k-1}{j.\binom{k}{j}} + k.\binom{k}{k} - k.\binom{k}{k} +\sum_{j=0}^{k-1}{\binom{k}{j}} =[/tex]

[tex]\color{red}\sum_{j=0}^{k}{j.\binom{k}{j}} - k + \sum_{j=0}^{k-1}{\binom{k}{j}}[/tex]


Entonces

[tex]k.2^{k-1} -{\color{red}\sum_{i=1}^{k}{i.\binom{k}{i-1}}} + (k+1)=[/tex]

[tex]k.2^{k-1} -{\color{red}\left[\sum_{j=0}^{k}{j.\binom{k}{j}} - k + \sum_{j=0}^{k-1}{\binom{k}{j}}\right]} + (k+1)=[/tex]

[tex]k.2^{k-1} -\sum_{j=0}^{k}{j.\binom{k}{j}} + k - \sum_{j=0}^{k-1}{\binom{k}{j}} + (k+1)=[/tex]

usando otra vez la HI

[tex]k.2^{k-1} -k.2^{k-1} + k - \sum_{j=0}^{k-1}{\binom{k}{j}} + (k+1)=[/tex]

[tex]k - \sum_{j=0}^{k-1}{\binom{k}{j}} + k+1=[/tex]

[tex]2k +1 - \sum_{j=0}^{k-1}{\binom{k}{j}} {\color{red}= (k+1)2^k}[/tex]

y tengo que probar esto ultimo...

Sea Q(k): [tex]\sum_{j=0}^{k-1}{\binom{k}{j}} = 2k +1 - (k+1)2^k[/tex]

*Q(1): 
[tex]\sum_{j=0}^{1-1}{\binom{1}{j}} = \binom{1}{0} = {\color{red}1}[/tex]
[tex]2.1 +1 - (1+1)2^1 = 3 - 4 = {\color{red}-1}[/tex]

*Q(2): 
[tex]\sum_{j=0}^{2-1}{\binom{2}{j}} = \binom{2}{0} + \binom{2}{1} = 1 + 2 = {\color{red}3}[/tex]
[tex]2.2 +1 - (2+1)2^2 = 5 - 3.4 = {\color{red}-7}[/tex]

*Q(3): 
[tex]\sum_{j=0}^{3-1}{\binom{3}{j}} = \binom{3}{0} + \binom{3}{1} + \binom{3}{2} = 1 + 3  + 3 = {\color{red}7}[/tex]
[tex]2.3 +1 - (3+1)2^3 = {\color{red}-25}[/tex]

No creo que se cumpla para ningun k natural, entonces que hice mal?


Que hice mal?