If, for a positive integer `n ,` the quadratic equation, `x(x+1)+(x-1)(x+2)++(x+ n-1)(x+n)=10 n` has two consecutive integral solutions, then `n` is equal to : ` (1) 10` (2) `11` (3) `12` (4) `9`

Options

A. 2
B. 3
C. 0
D. 11

Answer

D

Explanation

The given quadratic equation is `nx^(2) + x {1+3+5+...+(2n-1)}+{1*2+2*3+...+(n-1)n} = 10n `
Now, `1*2+2*3+...+(n-1)n = underset(r=1)overset(n-1)sum(r-1) r = underset(r=1)overset(n-1)sum r^(2) - underset(r=1)overset(n-1)sum r`
`" "=(n(n-1)(2n-1))/(6)-(n(n-1))/(2)=(n(n^(2)-1))/(3)`
So, the equation becomes `nx^(2) + n^(2)x + (n(n^(2)-1))/(3)-10n = 0`
or, `x^(2) + nx + ((n^(2)-31))/(3))=0`
Let `alpha, beta` be the roots of this equation. Then, `alpha + beta = - n and alpha beta = (n^(2)-31)/(3)`. It is given that `alpha and beta` are consexutive integers.
`therefore" "alpha-beta = +- 1`
`rArr" "(alpha-beta)^(2) = 1`
`rArr" "(alpha+beta)^(2)-4 alpha beta = 1`
`rArr" "n^(2)-(4(n^(2)-31))/(3) = 1`
`rArr" "3n^(2)-4n^(2) + 124 = 3 rArr n^(2) = 121 rArr n = 11`