If, for a positive integern, the quadratic equation, x(x+1)+(x+1)(x+2)+...+(x+n−1¯)(x+n)=10nhas two consecutive integral solutions, then n is equal to

Given Equation x(x+1)+(x+1)(x+2)+...+(x+n−1)(x+n)=10n⇒ nx2+x(1+3+5+...+2n−1)+(1.2+2.3+...+(n−1)n)−10n=0⇒ nx2+xn2+n(n2−1)3−10n=0⇒ x2+nx+n2−13−10=0Let α and β be the roots of the equation thenα+β=−n and αβ= (n2−1)3−10and (α−β)2=1⇒ (α+β)2−4αβ=1⇒ n2−4(n2−13−10)=1⇒ 3n2−4n2+4+120−3=0⇒ n2=121⇒ n= 11

If, for a positive integern, the quadratic equation, x(x+1)+(x+1)(x+2)+...+(x+n−1¯)(x+n)=10nhas two consecutive integral solutions, then n is equal to

Given Equation x(x+1)+(x+1)(x+2)+...+(x+n−1)(x+n)=10n⇒ nx2+x(1+3+5+...+2n−1)+(1.2+2.3+...+(n−1)n)−10n=0⇒ nx2+xn2+n(n2−1)3−10n=0⇒ x2+nx+n2−13−10=0Let α and β be the roots of the equation thenα+β=−n and αβ= (n2−1)3−10and (α−β)2=1⇒ (α+β)2−4αβ=1⇒ n2−4(n2−13−10)=1⇒ 3n2−4n2+4+120−3=0⇒ n2=121⇒ n= 11

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$\begin{array}{l} I f, f o r a p o s i t i v e int e g e r n, t h e q u a d r a t i c e q u a t i o n, x (x + 1) + (x + 1) (x + 2) + ... + (x + \bar{n - 1}) (x + n) = 10 n \\ h a s t w o c o n \sec u t i v e int e g r a l s o l u t i o n s, t h e n n i s e q u a l t o \end{array}$

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x (x + 1) + (x + 1) (x + 2) + ... + (x + n - 1) (x + n) = 10 n \\ \Rightarrow n x^{2} + x (1 + 3 + 5 + ... + 2 n - 1) + (1.2 + 2.3 + ... + (n - 1) n) - 10 n = 0 \\ \Rightarrow n x^{2} + x n^{2} + \frac{n (n^{2} - 1)}{3} - 10 n = 0 \\ \Rightarrow x^{2} + n x + \frac{n^{2} - 1}{3} - 10 = 0 \\ L e t α a n d β b e t h e r o o t s o f t h e e q u a t i o n t h e n \\ α + β = - n a n d α β = \frac{(n^{2} - 1)}{3} - 10 \\ a n d {(α - β)}^{2} = 1 \\ \Rightarrow {(α + β)}^{2} - 4 α β = 1 \\ \Rightarrow n^{2} - 4 (\frac{n^{2} - 1}{3} - 10) = 1 \\ \Rightarrow 3 n^{2} - 4 n^{2} + 4 + 120 - 3 = 0 \\ \Rightarrow n^{2} = 121 \\ \Rightarrow n = 11 \end{array}$