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16 Sep 2017, 09:55
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A
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64%
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Let \(x\) and \(y\) be the solutions of the equation \(a^2+3a-1=0\). If \(P_n\) \(= x^n+y^n\) for \(n≥0\), then, for \(n≥2\), \(P_n =\)
A) \(-3P_{n-1} + P_{n-2}\)
B) \(3P_{n-1} + P_{n-2}\)
C) \(-P_{n-1} + P_{n-2}\)
D) \(P_{n-1} + 3P_{n-2}\)
E) \(P_{n-1} + P_{n-2}\)
A) \(-3P_{n-1} + P_{n-2}\)
B) \(3P_{n-1} + P_{n-2}\)
C) \(-P_{n-1} + P_{n-2}\)
D) \(P_{n-1} + 3P_{n-2}\)
E) \(P_{n-1} + P_{n-2}\)
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I am not sure about shortcut method. But find below the method i used:
\(a^2+3a−1=0 => x+y = \frac{-3}{1} = -3\) and \(xy=\frac{-1}{1}=-1\) ( By sum and product of quadratic equation formula)
\(P_n =x^n+y^n\) \(n≥0\)
\(P_1 = x+y = -3\)
\(P_2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11\)
\(P_3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36\)
Now with respect to \(P_3\) =>\(P_2 =P_{n-1}\) and \(P_1=P_{n-2}\)
but putting values we got \(P_3 =-36 = -3*(11)+(-3) = -3*P_{n-1}+P_{n-2}\)
Answer: A
\(a^2+3a−1=0 => x+y = \frac{-3}{1} = -3\) and \(xy=\frac{-1}{1}=-1\) ( By sum and product of quadratic equation formula)
\(P_n =x^n+y^n\) \(n≥0\)
\(P_1 = x+y = -3\)
\(P_2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11\)
\(P_3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36\)
Now with respect to \(P_3\) =>\(P_2 =P_{n-1}\) and \(P_1=P_{n-2}\)
but putting values we got \(P_3 =-36 = -3*(11)+(-3) = -3*P_{n-1}+P_{n-2}\)
Answer: A
16 Sep 2017, 12:27
Kudos
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Nikkb
I am not sure about shortcut method. But find below the method i used:
\(a^2+3a−1=0 => x+y = \frac{-3}{1} = -3\) and \(xy=\frac{-1}{1}=-1\) ( By sum and product of quadratic equation formula)
\(Pn =x^n+y^n\) \(n≥0\)
\(p1 = x+y = -3\)
\(p2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11\)
\(p3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36\)
Now for p3 =>\(p2 =p(n-1)\) and \(p1=p(n-2)\)
but putting values we got \(p3 =-36 = -3*(11)+(-3) = -3*p(n-1)+p(n-1)\)
Answer: A
\(a^2+3a−1=0 => x+y = \frac{-3}{1} = -3\) and \(xy=\frac{-1}{1}=-1\) ( By sum and product of quadratic equation formula)
\(Pn =x^n+y^n\) \(n≥0\)
\(p1 = x+y = -3\)
\(p2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11\)
\(p3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36\)
Now for p3 =>\(p2 =p(n-1)\) and \(p1=p(n-2)\)
but putting values we got \(p3 =-36 = -3*(11)+(-3) = -3*p(n-1)+p(n-1)\)
Answer: A
Hi Nikkb,
We don't need a short-cut method here as the question can be solved in less than 2 minutes. Your approach is correct, however calculations could be made bit simpler to save time.
As you have rightly mentioned: Sum of roots, \(x+y = -3\) and Product of roots, \(xy=-1\)
Now we need to find \(P_n\) for \(n≥2\). This can be done by simply finding \(P_2\)
If you observe all answer choices have only \(P_{n-1}\) and \(P_{n-2}\). Hence we need to find \(P_1\) & \(P_0\)
\(P_0\) \(= x^0+y^0 = 1+1 =2\)
\(P_1\) \(= x^1+y^1 = x+y=-3\)
\(P_2\) \(=x^2+y^2 = (x+y)^2-2xy = (-3)^2-2*-1= -3*-3+2 =\) \(-3P_1\) \(+ P_0\)
Or \(P_n\) \(= −3P_{n−1}\) \(+ P_{n−2}\)
Hence Option A
