n∑k=1(x+k−1)(x+k)=10n
⇒n∑k=1[x2+(2k−1)x+(k−1)k]=10n
Now, n∑k=1(2k−1)=2n(n+1)2−n=n2
n∑k=1k(k−1)=n∑k=1k2−k=n(n+1)(2n+1)6−n(n+1)2=n(n2−1)3
So, nx2+n2x+n(n2−1)3−10n=0⇒3x2+3nx+(n2−31)=0 ⋯(1)
Now, given that the roots of equation (1) is α,α+1
Sum of roots
2α+1=−n⇒α=−n+12 ⋯(2)
Product of roots
α(α+1)=n2−313
Using equation (2),
−n+12(−n+12+1)=n2−313⇒n2−14=n2−313⇒n2=121∴n=11