Let , +2, Z, be the roots of the quadratic equation x(x+2)+(x+1)(x+3)+(x+2)(x+4)+ +(x+n-1)(x+n+1)=4n for some n N. Then n+ is equal to :
- 0
- 1
- 2
- 3
Answer
(3) 2
Solution
LHS = r=0n-1(x+r)(x+r+2)
=nx2+x r=0n-1(2r+2)+ r=0n-1r(r+2)
=nx2+n(n+1)x+ n(n-1)(2n+5)6
∴ Dividing the equation by n:
x2+(n+1)x+ (n-1)(2n+5)6-4=0, whose roots are and +2.
Sum of roots: 2 +2=-(n+1)
=- n+32 and +2=- n-12
Product of roots: ( +2)= (n+3)(n-1)4= (n-1)(2n+5)6-4
3(n+3)(n-1)=2(n-1)(2n+5)-48
3n2+6n-9=4n2+6n-58
n2=49 n=7 (∵ n )
∴ =- 7+32=-5 (roots -5 and -3)
n+ =7-5=2
Answer: (3)