Let =3+4+8+9+13+14+ upto 40 terms.

Question

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

Accepted Answer

1. A series sum is defined : $\alpha =3+4+8+9+13+14+\ldots ..{t}_{40}$

2. $(	an \beta {)}^{\frac{\alpha }{1020}}$ is a root of the quadratic equation ${x}^{2}+x-2=0$ where $\beta \in (0,\frac{\pi }{2})$

3. To evaluate the expression ${\sin }^{2}\beta +3{\cos }^{2}\beta $

$	herefore $ $\alpha =3+4+8+9+13+14+\ldots ..{t}_{40}$

$=7+17+27+\ldots ..+{T}_{20}$

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

$=(10)\cdot (14+190)=2,040\ldots \ldots (i)$

Equation (i) gives the value of $\alpha =2,040$ :

$(	an \beta {)}^{\frac{\alpha }{1020}}=(	an \beta {)}^{\frac{2040}{1020}}={	an }^{2}\beta \ldots \ldots (ii)$

Now let us solve the quadratic equation ${x}^{2}+x-2=0$ :

${x}^{2}+x-2=0\Rightarrow {x}^{2}+2x-x-2=0\Rightarrow (x+2)(x-1)=0$

$\Rightarrow \begin{gathered} p q \ -2 1 \end{gathered}$

where $p$ and $q$ are the roots

We have derived in equation $(ii)(	an \beta {)}^{\frac{\alpha }{1020}}={	an }^{2}\beta $ which is also a root of the equation ${x}^{2}+x-2=0$ :

From the pair of roots we get ${	an }^{2}\beta $ =1$\Rightarrow 	an \beta =\pm 1$ As ${	an }^{2}\beta \geqslant 0$ always

$\beta \in (0,\frac{\pi }{2})\Rightarrow 	an \beta =1\Rightarrow \beta =\frac{\pi }{4}$

$\Rightarrow {\sin }^{2}\beta +3{\cos }^{2}\beta ={\sin }^{2}\frac{\pi }{4}+3{\cos }^{2}\frac{\pi }{4}=(\frac{1}{2})+(\frac{3}{2})=2$

Solutions · 1