We have, x(x+1)+(x+1)(x+2)+...+(x+ul(n-1))(x+n),
=sum_(m=1)^(m=n){(x+ul(m-1))(x+m)},
=sum_(m=1)^(m=n){x^2+(2m-1)x+m(m-1)},
=x^2sum_(m=1)^(m=n)(1)+xsum_(m=1)^(m=n)(2m-1)+sum_(m=1)^(m=n){m(m-1)},
=x^2(n)+x{2sum_(m=1)^(m=n)m-sum_(m=1)^(m=n)1}+{sum_(m=1)^(m=n)m^2-sum_(m=1)^(m=n)m},
=nx^2+x{2*n/2(n+1)-n}+{n/6(n+1)(2n+1)-n/2(n+1)},
=nx^2+n^2x+n/6(n+1)(2n+1-3),
=nx^2+n^2x+n/3(n^2-1),
=n[x^2+nx+(n^2-1)/3].
Accordingly, the given quadr. eqn. becomes [because n ne 0],
x^2+nx+(n^2-1)/3=10, or, x^2+nx+(n^2-31)/3=0.
If alpha and beta are roots of this eqn., we have,
alpha+beta=-n, and alpha*beta=(n^2-31)/3.
:. (alpha-beta)^2=(alpha+beta)^2-4alpha*beta,
=n^2-4((n^2-31)/3),
=(124-n^2)/3.
But, given that, |alpha-beta|=1," we have, "1=(124-n^2)/3.
rArr n^2=121," giving, "n=+-11.
n in NN rArr n=11.
So, the right option is (3) 11.