Let α and β be the roots of x2-x-1=0, with α>β. For all positive integers n, definean=αn−βnα−β, n≥1b1=1 and bn=an−1+an+1, n≥2Then which of the following option(s) is/are correct?

α,β are roots of x2-x-1ar+2−ar=αr+2−βr+2−αr−βrα−β=αr+2−αr−βr+2−βrα−β=αrα2−1−βrβ2−1α−β=αrα−βrβα−β=αr+1−βr+1α−β=ar+1⇒ar+2−ar+1=ar⇒∑r=1nar=an+2−a2=an+2−α2−β2α−β=an+2−α+β=an+2−1now,∑n=1∞an10n=∑N=1∞α10n−∑N=1∞β10nα−βα101−α10−β101−β10α−β=α10−α−β10−βα−β=1010−α10−β=1089∑n=1∞bn10n=∑n=1∞an−1+an+110n=α101−α10+β101−β10=1289further bn=an−1+an+1=αn−1−βn−1+αn+1−βn+1α−βαβ=−1⇒αn−1=αnβ&βn−1=−αβn=αnα−β+α−ββnα−β=αn+βn

Let α and β be the roots of x2-x-1=0, with α>β. For all positive integers n, definean=αn−βnα−β, n≥1b1=1 and bn=an−1+an+1, n≥2Then which of the following option(s) is/are correct?

α,β are roots of x2-x-1ar+2−ar=αr+2−βr+2−αr−βrα−β=αr+2−αr−βr+2−βrα−β=αrα2−1−βrβ2−1α−β=αrα−βrβα−β=αr+1−βr+1α−β=ar+1⇒ar+2−ar+1=ar⇒∑r=1nar=an+2−a2=an+2−α2−β2α−β=an+2−α+β=an+2−1now,∑n=1∞an10n=∑N=1∞α10n−∑N=1∞β10nα−βα101−α10−β101−β10α−β=α10−α−β10−βα−β=1010−α10−β=1089∑n=1∞bn10n=∑n=1∞an−1+an+110n=α101−α10+β101−β10=1289further bn=an−1+an+1=αn−1−βn−1+αn+1−βn+1α−βαβ=−1⇒αn−1=αnβ&βn−1=−αβn=αnα−β+α−ββnα−β=αn+βn

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$Let α and β be the roots of x^{2} - x - 1 = 0, with α > β . For all positive integers$ n, define $a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1$ $b_{1} = 1 and b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$ Then which of the following option(s) is/are correct?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$b_{n} = α^{n} + β^{n} for all n \geq 1$

$a_{1} + a_{2} + a_{3} + \dots \dots . + a_{n} = a_{n + 2} - 1 for all n \geq 1$

$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \frac{12}{89}$

$\sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \frac{10}{89}$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$α, β are roots of x^{2} - x - 1$

$a_{r + 2} - a_{r} = \frac{(α^{r + 2} - β^{r + 2}) - (α^{r} - β^{r})}{α - β} = \frac{(α^{r + 2} - α^{r}) - (β^{r + 2} - β^{r})}{α - β}$

$= \frac{α^{r} (α^{2} - 1) - β^{r} (β^{2} - 1)}{α - β} = \frac{α^{r} α - β^{r} β}{α - β} = \frac{α^{r + 1} - β^{r + 1}}{α - β} = a_{r + 1}$

$\Rightarrow a_{r + 2} - a_{r + 1} = a_{r}$

$\Rightarrow \sum_{r = 1}^{n} a_{r} = a_{n + 2} - a_{2} = a_{n + 2} - \frac{α^{2} - β^{2}}{α - β} = a_{n + 2} - (α + β) = a_{n + 2} - 1$

now, $\sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \frac{\sum_{N = 1}^{\infty} {(\frac{α}{10})}^{n} - \sum_{N = 1}^{\infty} {(\frac{β}{10})}^{n}}{α - β}$

$\frac{\frac{\frac{α}{10}}{1 - \frac{α}{10}} - \frac{\frac{β}{10}}{1 - \frac{β}{10}}}{α - β} = \frac{\frac{α}{10 - α} - \frac{β}{10 - β}}{(α - β)} = \frac{10}{(10 - α) (10 - β)} = \frac{10}{89}$

$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \sum_{n = 1}^{\infty} \frac{a_{n - 1} + a_{n + 1}}{10^{n}} = \frac{\frac{α}{10}}{1 - \frac{α}{10}} + \frac{\frac{β}{10}}{1 - \frac{β}{10}} = \frac{12}{89}$