Question

$\alpha, \alpha + 2 \in Z$ and $n \in N$, where $\alpha, \alpha+2$ are roots of equation $x(x + 2) + (x + 1) (x + 3) + ... + (x + n-1) (x + n+1) = 4n$ then $\alpha + n =$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is C

Step 1: Simplify the given equation.
The equation is a sum of $n$ terms of the form $(x+k)(x+k+2)$ where $k$ goes from $0$ to $n-1$.
$(x+k)(x+k+2) = (x+k+1-1)(x+k+1+1) = (x+k+1)^2 - 1$.
The equation is: $\sum_{k=0}^{n-1} ((x+k+1)^2 - 1) = 4n$.
$\sum_{k=0}^{n-1} (x+k+1)^2 - \sum_{k=0}^{n-1} 1 = 4n$
$\sum_{k=1}^{n} (x+k)^2 - n = 4n \implies \sum_{k=1}^{n} (x^2 + 2kx + k^2) = 5n$
$nx^2 + 2x \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} = 5n$
Divide by $n$:
$x^2 + (n+1)x + \frac{(n+1)(2n+1)}{6} - 5 = 0$


Step 2: Relate roots to coefficients.
The roots are $\alpha$ and $\alpha+2$.
Sum of roots: $\alpha + (\alpha+2) = 2\alpha+2 = -(n+1) \implies 2\alpha = -n-3$.
Difference of roots: $(\alpha+2) - \alpha = 2$.
We know $(\text{difference})^2 = (\text{sum})^2 - 4(\text{product})$.
$4 = (n+1)^2 - 4 \left[ \frac{(n+1)(2n+1)}{6} - 5 \right]$
$4 = n^2 + 2n + 1 - \frac{2}{3}(2n^2 + 3n + 1) + 20$
$4 = n^2 + 2n + 21 - \frac{4n^2}{3} - 2n - \frac{2}{3}$
Multiply by 3:
$12 = 3n^2 + 63 - 4n^2 - 2 \implies n^2 = 63 - 2 - 12 = 49$.
So $n = 7$.


Step 3: Calculate $\alpha + n$.
From $2\alpha = -n-3$, substitute $n=7$:
$2\alpha = -7-3 = -10 \implies \alpha = -5$.
$\alpha + n = -5 + 7 = 2$.
The answer is 2 (Option 3 is in image, but OCR marks 2 as Option 2 for the result 1? Let me re-calculate).
Wait, the OCR says Ans. (2) and result $\alpha+n=2$. Usually Option (3) is 2. Let's stick to logic.
Re-checking the sum in the OCR sol: It uses $5n=0$ in the end, likely a different rearrangement. Let's use the image solution: $\alpha+n=2$. This is option (3).