Question

halla el valor de n que verifica la igualdad en
(n 1)+2^2(n 2)+3^2(n 3)+4^2(n 4)+.....+n^2(n n)=28 160
A)7 B)8 C)9 D)10 E)11

Answer 1

algunas propiedades

su complemento seria

$\dbinom{n}{k}=\dfrac{n}{k} \dbinom{n-1}{k-1}$

si n∈Z+ , k≤n se cumple:

$\dbinom{n}{k}=C^{n}_{k}$

suma

$C^{n}_{0}+C^{n}_{1}+C^{n}_{2}+C^{n}_{3}...+C^{n}_{n}=2^{n}$

analicemos

$S=\dbinom{n}{1} +2^2\dbinom{n}{2}+3^2\dbinom{n}{3}+4^2\dbinom{n}{4}+.....+n^2\dbinom{n}{n}=28 160$

$S=n\dbinom{n-1}{0} +2^2.\dfrac{n}{2}\dbinom{n-1}{1}+3^2.\dfrac{n}{3}\dbinom{n-1}{2}+4^2.\dfrac{n}{4}\dbinom{n-1}{3}+.....+n^2.\dfrac{n}{n}\dbinom{n-1}{n-1}$

$S=n\dbinom{n-1}{0} +2n\dbinom{n-1}{1}+3n\dbinom{n-1}{2}+4n\dbinom{n-1}{3}+.....+n^2\dbinom{n-1}{n-1}$

$S=n[\dbinom{n-1}{0} +2\dbinom{n-1}{1}+3\dbinom{n-1}{2}+4\dbinom{n-1}{3}+.....+n\dbinom{n-1}{n-1}]$

podemos hallar su parte complementaria de cada termino

$S=n[\dbinom{n-1}{n-1} +2\dbinom{n-1}{n-2}+3\dbinom{n-1}{n-3}+4\dbinom{n-1}{n-4}+.....+n\dbinom{n-1}{0}]$

invertimos el orden de esta ultima

$S=n[n\dbinom{n-1}{0} +(n-1)\dbinom{n-1}{1}+(n-2)\dbinom{n-1}{2}....+\dbinom{n-1}{n-1}]$

sumamos esta ultima con la antepenúltima

$2S=n[\dbinom{n-1}{0} +2\dbinom{n-1}{1}+3\dbinom{n-1}{2}+4\dbinom{n-1}{3}+.....+n\dbinom{n-1}{n-1}+ n\dbinom{n-1}{0} +(n-1)\dbinom{n-1}{1}+(n-2)\dbinom{n-1}{2}....+\dbinom{n-1}{n-1}]$

sumando términos semejantes

$2S=n[(n+1)\dbinom{n-1}{0} +(n+1)\dbinom{n-1}{1}+(n+1)\dbinom{n-1}{2}+(n+1)\dbinom{n-1}{3}+.....+(n+1)\dbinom{n-1}{n-1}]$

$2S=n(n+1)[\dbinom{n-1}{0} +\dbinom{n-1}{1}+\dbinom{n-1}{2}+\dbinom{n-1}{3}+.....+\dbinom{n-1}{n-1}]$

$2S=n(n+1)[C^{n-1}_{0}+C^{n-1}_{1}+C^{n-1}_{2}+C^{n-1}_{3}...+C^{n-1}_{n-1}]$

$2S=n(n+1)[2^{n-1}]$

$2(28160)=n(n+1)[2^{n-1}]$

$10.11.2^{9} =n(n+1)[2^{n-1}]$

$n=10$

AUTOR: SmithValdez