Let α = 3+4+8+9+13+14+ upto 40 terms. If ( β)^ (α)/(1020) is a root...

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

We can split the series into two arithmetic progressions, each containing $20$ terms:

$S_1 = 3 + 8 + 13 + \ldots$ up to $20$ terms

$S_2 = 4 + 9 + 14 + \ldots$ up to $20$ terms

Using the sum formula for an AP, $S_n = \dfrac{n}{2}[2a + (n-1)d]$:

$S_1 = \dfrac{20}{2}[2(3) + (20-1)5] = 10[6 + 95] = 1010$

$S_2 = \dfrac{20}{2}[2(4) + (20-1)5] = 10[8 + 95] = 1030$

Therefore, $\alpha = S_1 + S_2 = 1010 + 1030 = 2040$.

The exponent in the given expression is $\dfrac{\alpha}{1020} = \dfrac{2040}{1020} = 2$.

Thus, $(\tan\beta)^2 = \tan^2\beta$ is a root of the equation $x^2+x-2=0$.

Solving $x^2+x-2=0$ gives $(x+2)(x-1)=0 \Rightarrow x = -2$ or $x = 1$.

Since $\tan^2\beta \ge 0$, we must have $\tan^2\beta = 1$.

Given $\beta \in \left(0, \dfrac{\pi}{2}\right)$, $\tan\beta = 1 \Rightarrow \beta = \dfrac{\pi}{4}$.

We need to find the value of $\sin^2\beta + 3\cos^2\beta$.

Substituting $\beta = \dfrac{\pi}{4}$:

$\sin^2\left(\dfrac{\pi}{4}\right) + 3\cos^2\left(\dfrac{\pi}{4}\right) = \left(\dfrac{1}{\sqrt{2}}\right)^2 + 3\left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{1}{2} + \dfrac{3}{2} = 2$.

Answer: $2$