Question

Let \( \alpha, \beta ; \alpha>\beta \) be the roots of the eq. \( x^{2}-\sqrt{2} x-\sqrt{3}=0 \) Let \( P_{n}=\alpha^{n}-\beta^{n}, n \in N \), Then \( \left((11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}\right. \) is equal to :-

Answer 1

\(x^{2}-\sqrt{2 x}-\sqrt{3}=0\left\langle{ }_{\beta}^{\alpha}\right.\)

\(\alpha^{n+2}-\sqrt{2} \alpha^{n+1}-\sqrt{3} \alpha^{n}=0\)

and \(\beta^{n+2}-\sqrt{2} \beta^{n+1}-\sqrt{3} \beta^{n}=0\)

Subtracting

\( \left(\alpha^{n+2}-\beta^{n+2}\right)-\sqrt{2}\left(\alpha^{n+1}-\beta^{n+1}\right)-\sqrt{3}\left(\alpha^{n}-\beta^{n}\right)=0 \)

\( \Rightarrow P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_{n}=0\)

Put \(n=10\)

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

\( \mathrm{n}=9\)

\( \mathrm{P}_{11}-\sqrt{2} \mathrm{P}_{10}-\sqrt{3} \mathrm{P}_{9}=0 \)

\(11\left(\sqrt{3} \cdot \mathrm{P}_{10}+\sqrt{2} \mathrm{P}_{11}-\mathrm{P}_{11}\right)-10\left(\sqrt{2} \mathrm{P}_{10}-\mathrm{P}_{11}\right) \)

\(=0-10\left(-\sqrt{3} \mathrm{P}_{9}\right)=10 \sqrt{3} \mathrm{P}_{9}\)

Answer 2

Let α , β; α > β , ; > be the roots of the eq. x² − √ 2 x − √ 3 = 0  Let P_n = αⁿ − βⁿ, n ∈ N Then  ( 11√ 3 − 10√ 2 ) P₁₀ + ( 11√ 2 + 10 ) P₁₁ − 11P₁₂ = ?

Given α+β   =  √ 2  and  αβ= − √ 3 

− √ 3 = √ 2β  - β²  similarly  − √ 3 = √ 2α  -α ² 

( 11√ 3 − 10√ 2 ) P₁₀ + ( 11√ 2 + 10 ) P₁₁ − 11P₁₂ = 11√ 3P₁₀ + 11√ 2P₁₁ -11P₁₂ - 10√ 2P₁₀  + 10 P₁₁

= 11(√ 3P₁₀ + √ 2P₁₁ - P₁₂) - 10(√ 2P₁₀  - P₁₁)

= 11{ (√ 3(α¹⁰ − β¹⁰ ) + √ 2(α¹¹ − β¹¹ ) -  (α¹² − β¹² ) } -10{√ 2(α¹⁰ − β¹⁰ )  - (α¹¹ − β¹¹ ) }

= 11 { α¹⁰ [ √ 3 + √ 2α  - α² ] - β¹⁰[ √ 3 + √ 2β - β² ]  } - 10 { α⁹ [ √ 2α - α² ] - β⁹[ √ 2β - β² ] }

= 11 { α¹⁰ [ 0 ] - β¹⁰[ 0 ]  } - 10 { α⁹ [ -√ 3 ] - β⁹[ -√ 3 ] } = - 10 { - √ 3α⁹  +√ 3 β⁹ }

= 0 + 0 + 10√ 3{ α⁹ - β⁹ }  =  10√ 3P₉