If α=minimum of (x2+2x+3) and β=Ltn→∞∑r=1n1(r+2)r!, then ∑r=0nαrβn−2=

α=minimum of x2+2x+3=4.1.3−224=2β=Ltn→∞∑r=1n1(r+2)r!=Ltn→∞∑r=1nr+1(r+2)(r+1)r!=Ltn→∞∑r=1n(r+2)−1(r+2)!=Ltn→∞∑r=1n(1(r+1)!−1(r+2)!)=Ltn→∞[12−1(n+2)!]=12Now ∑r=0nαrβn−r=βn+αβn−1+α2βn−2+....+αn=βn[1+αβ+(αβ)2+.....+(αβ)n]=βn(αβ)n+1−1αβ−1=12n(4n+1−14−1)=4n+1−13.2n