If $\alpha$, $\beta$ are roots of the equation ${x^2} + 5(\sqrt 2 )x + 10 = 0$, $\alpha$ > $\beta$ and ${P_n} = {\alpha ^n} - {\beta ^n}$ for each positive integer n, then the value of $$\left( {{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}} \right)$$ is equal to _________.
Answer (integer)
1
Solution
${x^2} + 5\sqrt 2 x + 10 = 0$<br><br>& ${P_n} = {\alpha ^n} - {\beta ^n}$ (Given)<br><br>Now, $${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$$ = $${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$$<br><br>$${{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}$$<br><br>$${{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}$$<br><br>Since, $\alpha + 5\sqrt 2 = - 10/\alpha$ ..... (1)<br><br>& $\beta + 5\sqrt 2 = - 10/\beta$ ....... (2)<br><br>Now, put there values in above expression $= - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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