Question

calcular la suma de la serie
S=5Cn,0 /1+5^2Cn,1/2+5^3Cn,2/3+5^4Cn,3/4+....5^n+1Cn,n/n+1

Answer 1

el problema

$S=\dfrac{5C^{n}_{0}}{1}+\dfrac{5^2C^{n}_{1}}{2}+\dfrac{5^3C^{n}_{2}}{3}+\dfrac{5^4C^{n}_{3}}{4}+...+\dfrac{5^{n+1}C^{n}_{n}}{n+1}$

todos los términos tienen la forma

$\dfrac{5^{k+1}C^{n}_{k}}{k+1}.\dfrac{C^{n+1}_{k+1}}{\dfrac{n+1}{k+1} C^{n}_{k}}=\dfrac{5^{k+1}C^{n+1}_{k+1}}{n+1}$

entonces

$S=\dfrac{5^{1}C^{n+1}_{0+1}}{n+1}+\dfrac{5^{2}C^{n+1}_{1+1}}{n+1}+\dfrac{5^{3}C^{n+1}_{2+1}}{n+1}+...+\dfrac{5^{n+1}C^{n+1}_{n+1}}{n+1}$

$S=\dfrac{1}{n+1}[5^{1}C^{n+1}_{0+1}+5^{2}C^{n+1}_{1+1}+5^{3}C^{n+1}_{2+1}+...+5^{n+1}C^{n+1}_{n+1}]$

para que cumpla (1) sumemos lo que le falta

$S=\dfrac{1}{n+1}[5^{0}C^{n+1}_{0}+5^{1}C^{n+1}_{1}+5^{2}C^{n+1}_{2}+5^{3}C^{n+1}_{3}+...+5^{n+1}C^{n+1}_{n+1}-5^{0}C^{n+1}_{0}]$

la expresión dentro de la paréntesis tiene la forma de un binomio de newton

$(x+a)^{n}=\sum_{k=0}^nC^{n}_{k}.x^{n-k}.a^{k}$ (1)

entonces

$S=\dfrac{1}{n+1}[(1+5)^{n+1}-5^{0}C^{n+1}_{0}]$

$S=\dfrac{6^{n+1}-1}{n+1}$

AUTOR: SmithValdez