JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Quadratic Equations and Inequalities

a

and

b

are the roots of equation

x^{2} - 7 x - 1 = 0

, then the value of

\frac{a ^{21} + b ^{21} + a ^{17} + b ^{17}}{a ^{19} + b ^{19}}

is equal to

Answer: 51

Solution

Let

S_{n} = a^{n} + b^{n}

. Since

a

and

b

are roots of

x^{2} - 7 x - 1 = 0

, by Newton's recurrence relation (derived from the characteristic equation), we have:

S_{n + 2} = 7 S_{n + 1} + S_{n}

Applying this:

From the third equation:

S_{17} = S_{19} - 7 S_{18}

Substituting into the expression:

\frac{S _{21} + S _{17}}{S _{19}} = \frac{S _{21} + S _{19} - 7 S _{18}}{S _{19}}

From the second equation:

S_{18} = S_{20} - 7 S_{19}

Substituting:

\frac{S _{21} + S _{19} - 7 ( S _{20} - 7 S _{19} )}{S _{19}} = \frac{S _{21} + S _{19} - 7 S _{20} + 49 S _{19}}{S _{19}} = \frac{S _{21} - 7 S _{20} + 50 S _{19}}{S _{19}}

From the first equation:

S_{21} - 7 S_{20} = S_{19}

Therefore:

\frac{S _{19} + 50 S _{19}}{S _{19}} = \frac{51 S _{19}}{S _{19}} = 51

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer