The value of the positve integer `n` for which the quadratic equation `sum_(k=1)^n(x+k-1)(x+k)=10n` has solutions `alpha` and `alpha+1` for some `alpha` is

Options

A. 7
B. 11
C. 17
D. 25

Answer

B

Explanation

`:'sum_(k=1)^(n)(x+k-1)(x+k)=10n`
`impliessum_(k=1)^(n)x^(2)+x(2k-1)+(k-1)k=10n`
`impliesnx^(2)x(1+3+5+..+(2n-1))`
`+(0+1.2+2.3+3.4+……..+(n-1)n)=10n`
`impliesnx^(2)+x . n/2 (1+2n-1)`
`+((n(+1)(2n+1))/6-(n(n+1))/2)=10n`
`impliesnx^(2)+n^(2)x+(n(n^(2)-1))/3=10n`
`impliesx^(2)+nx+((n^(2)-31))/3=0` [dividing by `n`]
`:' (alpha +1)-alpha=(sqrt(D))/1`
`1=sqrt(D)`
`impliesD=1`
`impliesn^(2)-4.1. ((n^(2)-31))/3=1`
`implies3n^(2)-4c^(2)+124=3`
`impliesn^(2)=121`
`:.n=11`