Question

Find the value of positive integer ’ n’ for which the quadratic equation,n∑k=1(x+k−1)(x+k)=10n, has solutions α and α+1 for some α1210911

Solution

The correct option is D 11n∑k=1(x+k−1)(x+k)=10n∑[x2+(2k−1)x+(k−1)k]=10n n.x2+n2x+(n−1)n(n+1)3−10n=0 Let roots be α and β α+β=n αβ=(n−1)(n+1)3−10 Since, difference of root is 1, So,(α+β)2−4αβ=1 n2−4((n−1)(n+1)3)=13n2−4(n2−1−30)=33n2−4n2+124=3+n2=121n=±11n=11

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