Question

If, for a positive integer n, the quadratic equation, $x(x+1)+(x+1)(x+2)+....+(x + \overline{ n - 1}) (x+ n)=10n$ has two consecutive integral solutions, then $n$ is equal to :

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The Correct Option is C

To solve the given quadratic equation problem, let's start by understanding and expanding the sequence in the equation:

The given quadratic expression is:

S = x(x+1) + (x+1)(x+2) + \cdots + (x+n-1)(x+n)

We can express each term (x+k)(x+k+1) as:

(x+k)(x+k+1) = x^2 + 2kx + k^2 + k

This simplifies to:

x^2 + (2k+1)x + (k^2+k)

Let's sum up these terms from k=0 to n-1:

S = \sum_{k=0}^{n-1} \left( x^2 + (2k+1)x + (k^2+k) \right)

This turns into:

• x^2\sum_{k=0}^{n-1} 1 = nx^2
• x\sum_{k=0}^{n-1} (2k+1) = x(n^2) (because the sum of first n odd numbers is n^2)
• \sum_{k=0}^{n-1} (k^2+k) = \frac{(n-1)n(2n-1)}{6} + \frac{n(n-1)}{2}

Therefore, the sum S becomes:

S = nx^2 + n^2x + \frac{(n-1)n(2n-1)}{6} + \frac{n(n-1)}{2}

We are given:

S = 10n

Which means:

nx^2 + n^2x + \frac{(n-1)n(2n-1)}{6} + \frac{n(n-1)}{2} = 10n

Simplifying and removing n from terms where possible (for non-zero n):

x^2 + nx + \frac{(2n-1)(n-1)}{6} + \frac{n-1}{2} = 10

This is a quadratic equation in x:

x^2 + nx + \left(\frac{n^3 - n + 3n - 3}{6}\right) = 10

Simplify this equation further and note that it should have two consecutive integer solutions:

By solving, you find that n = 11 provides integer solutions.

Therefore, the correct answer is 11.