# The value of n=?

Mar 2, 2018

The right option is $\left(3\right) 11$.

#### Explanation:

We have, $x \left(x + 1\right) + \left(x + 1\right) \left(x + 2\right) + \ldots + \left(x + \underline{n - 1}\right) \left(x + n\right)$,

$= {\sum}_{m = 1}^{m = n} \left\{\left(x + \underline{m - 1}\right) \left(x + m\right)\right\}$,

$= {\sum}_{m = 1}^{m = n} \left\{{x}^{2} + \left(2 m - 1\right) x + m \left(m - 1\right)\right\}$,

$= {x}^{2} {\sum}_{m = 1}^{m = n} \left(1\right) + x {\sum}_{m = 1}^{m = n} \left(2 m - 1\right) + {\sum}_{m = 1}^{m = n} \left\{m \left(m - 1\right)\right\}$,

$= {x}^{2} \left(n\right) + x \left\{2 {\sum}_{m = 1}^{m = n} m - {\sum}_{m = 1}^{m = n} 1\right\} + \left\{{\sum}_{m = 1}^{m = n} {m}^{2} - {\sum}_{m = 1}^{m = n} m\right\}$,

$= n {x}^{2} + x \left\{2 \cdot \frac{n}{2} \left(n + 1\right) - n\right\} + \left\{\frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) - \frac{n}{2} \left(n + 1\right)\right\}$,

$= n {x}^{2} + {n}^{2} x + \frac{n}{6} \left(n + 1\right) \left(2 n + 1 - 3\right)$,

$= n {x}^{2} + {n}^{2} x + \frac{n}{3} \left({n}^{2} - 1\right)$,

$= n \left[{x}^{2} + n x + \frac{{n}^{2} - 1}{3}\right]$.

Accordingly, the given quadr. eqn. becomes [because n ne 0],

${x}^{2} + n x + \frac{{n}^{2} - 1}{3} = 10 , \mathmr{and} , {x}^{2} + n x + \frac{{n}^{2} - 31}{3} = 0$.

If $\alpha \mathmr{and} \beta$ are roots of this eqn., we have,

$\alpha + \beta = - n , \mathmr{and} \alpha \cdot \beta = \frac{{n}^{2} - 31}{3}$.

$\therefore {\left(\alpha - \beta\right)}^{2} = {\left(\alpha + \beta\right)}^{2} - 4 \alpha \cdot \beta$,

$= {n}^{2} - 4 \left(\frac{{n}^{2} - 31}{3}\right)$,

$= \frac{124 - {n}^{2}}{3}$.

But, given that, $| \alpha - \beta | = 1 , \text{ we have, } 1 = \frac{124 - {n}^{2}}{3}$.

$\Rightarrow {n}^{2} = 121 , \text{ giving, } n = \pm 11$.

$n \in \mathbb{N} \Rightarrow n = 11$.

So, the right option is $\left(3\right) 11$.
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