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#.
