JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 23)
If $$a$$ and $$b$$ are the roots of the equation $$x^{2}-7 x-1=0$$, then the value of $$\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$$ is equal to _____________.
Answer
51
Explanation
We have, $a$ and $b$ are the roots of the equation
$$ \begin{aligned} & x^2-7 x-1=0 \\\\ & \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i) \end{aligned} $$
On squaring both sides, we get $a^4+1=51 a^2$
Similarly, $b^4+1=51 b^2$ ...........(ii)
$$ \text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}} $$
$$ \begin{aligned} & =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\ & =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51 \end{aligned} $$
$$ \begin{aligned} & x^2-7 x-1=0 \\\\ & \Rightarrow a^2-7 a-1=0 \Rightarrow a^2-1=7 a .........(i) \end{aligned} $$
On squaring both sides, we get $a^4+1=51 a^2$
Similarly, $b^4+1=51 b^2$ ...........(ii)
$$ \text { Now, } \frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}=\frac{a^{17}\left(a^4+1\right)+b^{17}\left(b^4+1\right)}{a^{19}+b^{19}} $$
$$ \begin{aligned} & =\frac{a^{17}\left(51 a^2\right)+b^{17}\left(51 b^2\right)}{a^{19}+b^{19}} \quad[\because \text { From Eq. (i) and (ii) }] \\\\ & =\frac{51\left[a^{19}+b^{19}\right]}{a^{19}+b^{19}}=51 \end{aligned} $$
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