?

Debes descomponer la suma en dos partes:

S1=∑k=1n(n−1k−1)32(n−k)=∑k=1n−1(n−1k−1)32(n−k)+1S1=∑k=1n(n−1k−1)32(n−k)=∑k=1n−1(n−1k−1)32(n−k)+1

S2=−∑k=1n(n−1k−1)(−1)kk=−∑k=1n−1(n−1k−1)(−1)kk−(−1)nnS2=−∑k=1n(n−1k−1)(−1)kk=−∑k=1n−1(n−1k−1)(−1)kk−(−1)nn

La primera parte es sencilla si tienes en cuenta la fórmula del binomio de Newton:

(1+a)m=∑k=0m(mk)am−k(1+a)m=∑k=0m(mk)am−k

Es sencillo ver que si se toma a=32a=32 y m=n−1m=n−1 tenemos que:

(1+32)n−1=∑k′=1n−1(n−1k′−1)(32)(n−1)−(k′−1)+1(1+32)n−1=∑k′=1n−1(n−1k′−1)(32)(n−1)−(k′−1)+1

donde se ha definido k′=k+1k′=k+1:

10n−1=∑k=1n−1(n−1k−1)(32)n−k+1⇒S1=10n−110n−1=∑k=1n−1(n−1k−1)(32)n−k+1⇒S1=10n−1

Para la segunda parte se define la función:

(x−1)n−1=∑k=0n−1(n−1k)xk(−1)n−1−k(x−1)n−1=∑k=0n−1(n−1k)xk(−1)n−1−k

Y ahora se integra entre 0 y 1:

∫10(x−1)n−1 dx=[∑k=0n−1(n−1k)xk+1k+1(−1)n−1−k]10⇒∫01(x−1)n−1 dx=[∑k=0n−1(n−1k)xk+1k+1(−1)n−1−k]01⇒

(−1)n−1n=(−1)n−1∑k=0n−1(n−1k)(−1)kk+1⇒(−1)n−1n=(−1)n−1∑k=0n−1(n−1k)(−1)kk+1⇒

1n=−[∑k=1n−1(n−1k−1)(−1)k−1k−(−1)n−1n]⇒S2=−1n1n=−[∑k=1n−1(n−1k−1)(−1)k−1k−(−1)n−1n]⇒S2=−1n

Por lo tanto la suma buscada es:

S1+S2=10n−1−1nS1+S2=10n−1−1n