Solving by the quadratic formula, n=2−4±16+437⋅4=2−4±42=238=19 (since clearly n≥0 ). The answer is therefore 1+9=(C)10 .
OR
Dividing both sides by n ! gives
(n+1)+(n+2)(n+1)=440⇒n2+4n−437=0⇒(n−19)(n+23)=0
Since n is non-negative, n=19. The answer is 1+9=(C)10 .
OR
Dividing both sides by n ! as before gives (n+1)+(n+1)(n+2)=440. Now factor out (n+1), giving (n+1)(n+3)=440. By considering the prime factorization of 440 , a bit of experimentation gives us n+1=20 and n+3=22, so n=19, so the answer is 1+9=(C)10.
OR
Since (n+1)!+(n+2)!=(n+1)n!+(n+2)(n+1)n!=440⋅n!, the result can be factored into (n+1)(n+3)n!=440⋅n! and divided by n ! on both sides to get (n+1)(n+3)=440. From there, it is easier to complete the square with the quadratic (n+1)(n+3)=n2+4n+3, so n2+4n+4=441⇒(n+2)2=441. Solving for n results in n=19,−23, and since n>0,n=19 and the answer is 1+9=(C)10 .