If a and b are the roots of the equation x^2- | vedclass

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(B) Let $S_n = a^n + b^n$. Since $a$ and $b$ are roots of $x^2-7x-1=0$,by Newton's Sums,we have $S_{n+2} - 7S_{n+1} - S_n = 0$,which implies $S_{n+2}

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(B) Let $S_n = a^n + b^n$. Since $a$ and $b$ are roots of $x^2-7x-1=0$,by Newton's Sums,we have $S_{n+2} - 7S_{n+1} - S_n = 0$,which implies $S_{n+2} = 7S_{n+1} + S_n$.We want to evaluate $\frac{S_{21} + S_{17}}{S_{19}}$.From the recurrence relation,$S_{21} = 7S_{20} + S_{19}$.Also,$S_{19} = 7S_{18} + S_{17}$,which implies $S_{17} = S_{19} - 7S_{18}$.Substituting these into the expression:$\frac{S_{21} + S_{17}}{S_{19}} = \frac{7S_{20} + S_{19} + S_{19} - 7S_{18}}{S_{19}} = \frac{7S_{20} + 2S_{19} - 7S_{18}}{S_{19}}$.Since $S_{20} = 7S_{19} + S_{18}$,we have $S_{20} - S_{18} = 7S_{19}$.Substituting this into the numerator:$\frac{7(S_{20} - S_{18}) + 2S_{19}}{S_{19}} = \frac{7(7S_{19}) + 2S_{19}}{S_{19}} = \frac{49S_{19} + 2S_{19}}{S_{19}} = \frac{51S_{19}}{S_{19}} = 51$.

If $a$ and $b$ are the roots of the equation $x^2-7x-1=0$,then the value of $\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$ is equal to $........$.

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