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:
\(\sin^2\beta + 3\cos^2\beta = \frac{1}{2} + 3\left(\frac{1}{2}\right)\)
\(= \frac{1}{2} + \frac{3}{2}\)
\(= \frac{4}{2} = 2\)
Similar Questions
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
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- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
