If a and b are the roots of the equation x^2-7x-1=0, then the value of a^21+b^21+a^17+b^17…

Question

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

Answer: 51

Accepted Answer

Let $S_n=a^n+b^n$. Since $a,b$ satisfy $x^2=7x+1$, Newton's identity gives $S_{n+2}=7S_{n+1}+S_n$.

So $S_{21}=7S_{20}+S_{19}$ and $S_{19}=7S_{18}+S_{17} \implies S_{17}=S_{19}-7S_{18}$.

$$\frac{S_{21}+S_{17}}{S_{19}}=\frac{(7S_{20}+S_{19})+(S_{19}-7S_{18})}{S_{19}}=\frac{7(S_{20}-S_{18})+2S_{19}}{S_{19}}.$$

Since $S_{20}=7S_{19}+S_{18}$, we get $S_{20}-S_{18}=7S_{19}$, so:
$$\frac{7(7S_{19})+2S_{19}}{S_{19}}=\frac{49S_{19}+2S_{19}}{S_{19}}=51.$$

Solutions · 1