Let $\alpha$ and $\beta$ be the roots of $x^{2}-x-1=0$, with $\alpha>\beta$. For all positive integers n, define $$\begin{aligned} & a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}, n \geq 1 \\ & b_{1}=1 \text { and } b_{n}=a_{n-1}+a_{n+1}, n \geq 2 . \end{aligned}$$ Then which of the following options is/are correct ?

(A) $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+$..... $+\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}+2}-1$ for all $\mathrm{n} \geq 1$; (B) $\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{10^{\mathrm{n}}}=\frac{10}{89}$; (D) $\mathrm{b}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$ for all $\mathrm{n} \geq 1$

(C) $\sum_{\mathrm{n}=1}^{\infty} \frac{\mathrm{b}_{\mathrm{n}}}{10^{\mathrm{n}}}=\frac{8}{89}$

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Sequences and Series

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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 options is/are correct ?

(A) $a_{1} + a_{2} + a_{3} +$ ..... $+ a_{n} = a_{n + 2} - 1$ for all $n \geq 1$ ✓ Correct
(B) $\sum_{n = 1}^{\infty} \frac{a _{n}}{1 0 ^{n}} = \frac{10}{89}$ ✓ Correct
(C) $\sum_{n = 1}^{\infty} \frac{b _{n}}{1 0 ^{n}} = \frac{8}{89}$
(D) $b_{n} = α^{n} + β^{n}$ for all $n \geq 1$ ✓ Correct

Answer: (A), (B), (D)

Solution

Since

α, β

are roots of

x^{2} - x - 1 = 0

, we have

α^{2} = α + 1

and

β^{2} = β + 1

.

For the recurrence:

a_{r + 2} - a_{r + 1} = a_{r}

(derived from the characteristic equation).

Summing:

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

✓

For the infinite sum:

n = 1 \sum \infty \frac{a _{n}}{1 0 ^{n}} = \frac{10}{( 10 - α ) ( 10 - β )} = \frac{10}{100 - 10 - 1} = \frac{10}{89}

✓

For

b_{n}

: Since

α β = - 1

b_{n} = a_{n - 1} + a_{n + 1} = \frac{α ^{n} ( α - β ) + β ^{n} ( α - β )}{α - β} = α^{n} + β^{n}

✓

For option (C), the calculation would give a different value.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Sequences and Series

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