Question

Given n =1 + x and x is the product of four consecutive integers. Then n will be

Answer 1

Let the $4$ no be $a;(a+1);(a+2);(a+3)$

As product of 4 consecutive interference is even

$\Rightarrow$ x is even $\therefore 1+xisodd$

$x=a(a+1)(a+2)(a+3)=a^4+6a^3+11a^2+6a$

$\Rightarrow n=1+x=a^4+6a^3+11a^2+6a+1$

$=(a^2+3a+1{)}^2$

$\therefore {}^{\prime }{n}^{\prime }isalsoaperfectsquare$