Let, be the roots of the equation x 2 - 2 x - 3 = 0 . Let. Then ( 11 3 - 10 2 ) P 10 + ( 11 2 + 10 ) P 11 - 11 P 12 is equal to [2024]

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Solution

Q.
Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

1 $11 \sqrt{2} P_{9}$

2 $10 \sqrt{2} P_{9}$

3 $10 \sqrt{3} P_{9}$

4 $11 \sqrt{3} P_{9}$

Ans.
(3)

We have, $x^{2} - \sqrt{2} x - \sqrt{3} = 0$ has two roots $α$ and $β$

So for $S_{n} = α^{n} \neq β^{n},$ by Newton's theorem, we have $S_{n + 2} - \sqrt{2} S_{n + 1} - \sqrt{3} S_{n} = 0$

$\Rightarrow (α^{n + 2} - β^{n + 2}) - \sqrt{2} (α^{n + 1} - β^{n + 1}) - \sqrt{3} (α^{n} - β^{n}) = 0$

$\Rightarrow P_{n + 2} - \sqrt{2} P_{n + 1} - \sqrt{3} P_{n} = 0 [∵ P_{n} = α^{n} - β^{n}]$

For $n = 9,$ we have

$P_{12} - \sqrt{2} P_{11} - \sqrt{3} P_{10} = 0$ ...(i)

$\Rightarrow P_{12} = \sqrt{2} P_{11} + \sqrt{3} P_{10}$

For $n = 9,$ we have

$P_{11} - \sqrt{2} P_{10} - \sqrt{3} P_{9} = 0$

$\Rightarrow P_{11} = \sqrt{2} P_{10} + \sqrt{3} P_{9}$

So, $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$

$= (11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) (\sqrt{2} P_{10} + \sqrt{3} P_{9}) - 11 (\sqrt{2} P_{11} + \sqrt{3} P_{10})$

$= 11 \sqrt{3} P_{10} - 10 \sqrt{2} P_{10} + 22 P_{10} + 10 \sqrt{2} P_{10} - 11 \sqrt{3} P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 11 \sqrt{2} (\sqrt{2} P_{10} + \sqrt{3} P_{9})$

$= 22 P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 22 P_{10} - 11 \sqrt{6} P_{9} = 10 \sqrt{3} P_{9}$

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