If - are roots of the equation x2-5-2x-10-0- - - and Pn-n-n for each positive integer n- then the value of P17P20-5-2P17P19P18P19-5-2P218 is equal to

Α^n 2(α^2+5√(2)α+10)=0 …(1) β^n 2(β^2+5√(2)β+10)=0 …(2) From (2) (1) P_n+5√(2)P_n 1= 10P_n 2 Now, nbsp; P_17(P_20+5√(2)P_19)P_18(P_19+ 5√(2)P_18)=P_17.( 1 ...

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Standard XII

Mathematics

Properties of Iota

If α,β are ro...

Question

If $α, β$ are roots of the equation $x^{2} + 5 (\sqrt{2}) x + 10 = 0,$ $α > β$ and $P_{n} = α^{n} - β^{n}$ for each positive integer $n$ , then the value of $(\frac{P_{17} P_{20} + 5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19} + 5 \sqrt{2} P_{18}^{2}})$ is equal to

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Solution

$α^{n - 2} (α^{2} + 5 \sqrt{2} α + 10) = 0 \dots (1)$
$β^{n - 2} (β^{2} + 5 \sqrt{2} β + 10) = 0 \dots (2)$
From $(2) - (1)$
$P_{n} + 5 \sqrt{2} P_{n - 1} = - 10 P_{n - 2}$
Now,
$\frac{P_{17} (P_{20} + 5 \sqrt{2} P_{19})}{P_{18} (P_{19} + 5 \sqrt{2} P_{18})} = \frac{P_{17} . (- 10 P_{18})}{P_{18} . (- 10 P_{17})} = 1$

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If α,β are roots of the equation x2+5(√2)x+10=0, α>β and Pn=αn−βn for each positive integer n, then the value of (P17P20+5√2P17P19P18P19+5√2P218) is equal to

αn−2(α2+5√2α+10)=0 …(1) βn−2(β2+5√2β+10)=0 …(2) From (2)−(1) Pn+5√2Pn−1=−10Pn−2 Now, P17(P20+5√2P19)P18(P19+5√2P18)=P17.(−10P18)P18.(−10P17)=1

If $α, β$ are roots of the equation $x^{2} + 5 (\sqrt{2}) x + 10 = 0,$ $α > β$ and $P_{n} = α^{n} - β^{n}$ for each positive integer $n$ , then the value of $(\frac{P_{17} P_{20} + 5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19} + 5 \sqrt{2} P_{18}^{2}})$ is equal to