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Question

The value of positive integer n for which the quadratic equation, nk=1(x+k1)(x+k)=10n has solutions α and α+1 for some α, is

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Solution

nk=1(x+k1)(x+k)=10n
nk=1[x2+(2k1)x+(k1)k]=10n
Now, nk=1(2k1)=2n(n+1)2n=n2
nk=1k(k1)=nk=1k2k=n(n+1)(2n+1)6n(n+1)2=n(n21)3
So, nx2+n2x+n(n21)310n=03x2+3nx+(n231)=0 (1)

Now, given that the roots of equation (1) is α,α+1
Sum of roots
2α+1=nα=n+12 (2)
Product of roots
α(α+1)=n2313
Using equation (2),
n+12(n+12+1)=n2313n214=n2313n2=121n=11

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