The value of n=?

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1 Answer
Mar 2, 2018

The right option is # (3) 11#.

Explanation:

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#.