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. 11
B. 12
C. 9
D. 10

Answer

1

Explanation

We have
`x (x + 1) + (x + 1)(x + 2) + …+ (x + bar(n-1)) (x + n) = 10 n`
Roots of above equation are consecutive integers.
Let roots be ` alpha and alpha + 1 `.
`rArr alpha ( alpha + 1) + (alpha + 1)(alpha + 2) + …+ (alpha + bar(n-1)) (alpha + n) = 10 n` (2)
Subtracting (2) from (1) , we get
` alpha (alpha + 1)-(alpha + n) (alpha + n + 1) = 0 `
`rArr alpha ^(2) + alpha - alpha^(2) - (2n + 1) alpha - n(n+ 1) = 0 `
`rArr alpha = - (n+1)/(2)`
Putting this value in the original equation , i.e.,
`nx^(2)sum_(r=1)^(n) (2r -1) + sum _(r=1)^(n) (r - 1) r = 10n ` , we get
`( 3(n+ 1)^(2))/(4) - ((n+1))/(2) n^(2) + (n (n+1) (2n+1))/(6) - (n(m+1))/(2) = 10 n`
`rArr 3(n+ 1)^(2)- 6n(n+1) + 2(n+1) (2n +1) - 6(n+1) = 120`
`rArr n^(2) = 121`
`rArr n = 11`