Question
If (n + 3)! = 56 [(n + 1)!], find n.
(n + 3)! = 56 [(n + 1)!]
⇒ (n + 3)! × (n + 2)! × (n + 1)! = 56 [(n - 1)!] ⇒ (n + 2)(n + 3) = 56 ⇒ n2 + 3n + 2n + 6 = 56 ⇒ n2 + 5n + 6 - 56 = 0 ⇒ n2 + 5n + 50 = 0 ⇒ n2 + 10n - 5n - 50 = 0 ⇒ n (n + 10) - 5 (n + 10) = 0 ⇒ (n - 5)(n + 10) = 0 $\big[\therefore \text{n} +10 \neq 0\big]$ ⇒ n - 5 = 0 ⇒ n - 5 = 0 ⇒ n = 5 Hence, n = 5Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Using binomial theorem write down the expansions of the following:
$(^3\sqrt{\text{x}}-^3\sqrt{\text{a}})^6$