Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2 + x - 2 = 0$, $\beta \in \left(0, \frac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to :

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Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2 + x - 2 = 0$, $\beta \in \left(0, \frac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to :

$\frac{7}{4}$

$\frac{5}{2}$

$\frac{3}{2}$

The correct answer is

Step 1: Calculate the sum $\alpha$

The series is given by $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ up to 40 terms.

Group the terms into pairs: $(3+4) + (8+9) + (13+14) + \dots$. This gives 20 pairs.

The sums of these pairs form a new sequence: $7, 17, 27, \dots$.

This sequence is an arithmetic progression (AP) with the first term $A_1 = 7$ and the common difference $D = 10$. There are $n = 20$ terms in this AP.

Use the AP sum formula $S_n = \frac{n}{2}(2A_1 + (n-1)D)$.

$\alpha = S_{20} = \frac{20}{2}(2(7) + (20-1)10)$

$\alpha = 10(14 + 19 \times 10)$

$\alpha = 10(14 + 190)$

$\alpha = 10(204) = 2040$

Step 2: Find the roots of the quadratic equation

The equation is $x^2 + x - 2 = 0$.

Factor the equation: $(x+2)(x-1) = 0$.

The roots are $x = 1$ and $x = -2$.

Step 3: Determine the value of $(\tan\beta)^2$

We are given that $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root.

Calculate the exponent: $\frac{\alpha}{1020} = \frac{2040}{1020} = 2$.

Therefore, $(\tan\beta)^2$ is a root of the equation.

This means $(\tan\beta)^2 = 1$ or $(\tan\beta)^2 = -2$.

Step 4: Determine $\tan\beta$

Given $\beta \in \left(0, \frac{\pi}{2}\right)$, $\tan\beta$ must be positive.

The possibility $(\tan\beta)^2 = -2$ is discarded as squares of real numbers are non-negative.

So, $(\tan\beta)^2 = 1$.

Since $\tan\beta > 0$, we have $\tan\beta = \sqrt{1} = 1$.

Step 5: Find $\beta$ and evaluate the expression

For $\tan\beta = 1$ and $\beta \in \left(0, \frac{\pi}{2}\right)$, we have $\beta = \frac{\pi}{4}$.

We need to evaluate $\sin^2\beta + 3\cos^2\beta$.

At $\beta = \frac{\pi}{4}$, $\sin\beta = \frac{1}{\sqrt{2}}$ and $\cos\beta = \frac{1}{\sqrt{2}}$.

Then, $\sin^2\beta = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$ and $\cos^2\beta = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

Substitute these values:

$\sin^2\beta + 3\cos^2\beta = \frac{1}{2} + 3\left(\frac{1}{2}\right)$

$= \frac{1}{2} + \frac{3}{2}$

$= \frac{4}{2} = 2$

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