Let , be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) +...(x + n – 1)(x + n + 1) = 4n for some . Then is equal to [2026]

Question

Let $α, α + 2, α \in Z$ , be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) +...(x + n – 1)(x + n + 1) = 4n for some $n \in Z$ . Then $n + α$ is equal to [2026]

Smart Achievers · Accepted Answer

(3)

$\sum_{n = 1}^{n} (x + n - 1) (x + n + 1) = 4 n$

$\sum_{n = 1}^{n} [x^{2} + x (2 n) + (n^{2} - 1)] = 4 n$

$n x^{2} + 2 x \frac{n (n + 1)}{2} + \frac{n (n + 1) (2 n + 1)}{6} = 5 n$

$x^{2} + (n + 1) x + \frac{(n + 1) (2 n + 1) - 30}{6} = 0$

Roots of above equation are $α$ and $α + 2$

Sum of roots i.e., $α + α + 2 = - n - 1$

$\Rightarrow 2 α + n = - 3$ ... (i)

Products of roots i.e., $α (α + 2) = \frac{(n + 1) (2 n + 1) - 30}{6}$

$\Rightarrow 6 α^{2} + 12 α = (- 3 - 2 α + 1) (- 6 - 4 α + 1) - 30$ [Using (i)]

$\Rightarrow 6 α^{2} + 12 α = (2 α + 2) (4 α + 5) - 30$

$\Rightarrow 6 α^{2} + 12 α = 8 α^{2} + 18 α + 10 - 30$

$\Rightarrow 2 α^{2} + 6 α - 20 = 0$

$\Rightarrow α^{2} + 3 α - 10 = 0$

$\Rightarrow (α + 5) (α - 2) = 0$

$\Rightarrow α = 2 or α = - 5$

$∴$ From (i)

If $α = 2$ , then n = 7 and if $α = - 5$ , then n = –7 [Not possible]

$∴ α + n = - 5 + 7 = 2$

Solution

Q.
Let $α, α + 2, α \in Z$ , be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) +...(x + n – 1)(x + n + 1) = 4n for some $n \in Z$ . Then $n + α$ is equal to [2026]

1 0

2 1

3 2

4 3

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Solution

Q. Let α, α+2, α∈Z, be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) +...(x + n – 1)(x + n + 1) = 4n for some n∈Z. Then n+α is equal to [2026]

1 0

2 1

3 2

4 3

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Q.
Let $α, α + 2, α \in Z$ , be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) +...(x + n – 1)(x + n + 1) = 4n for some $n \in Z$ . Then $n + α$ is equal to [2026]