JEE Advanced 2019 · Paper 1 · Mathematics Q5

← JEE Advanced 2019 · Paper 1

MathematicsMultiple correct+4 / -1Question 41 of 54

Let $\alpha$ and $\beta$ be the roots of $x^2 - x - 1 = 0$ , with $\alpha > \beta$ . For all positive integers $n$ , define

$a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta},\quad n \ge 1,$

$b_1 = 1\quad\text{and}\quad b_n = a_{n-1} + a_{n+1},\quad n \ge 2.$

Then which of the following options is/are correct?

a_1 + a_2 + a_3 + \cdots + a_n = a_{n+2} - 1

for all

n \ge 1

\displaystyle\sum_{n=1}^\infty\frac{a_n}{10^n} = \frac{10}{89}

b_n = \alpha^n + \beta^n

for all

n \ge 1

\displaystyle\sum_{n=1}^\infty\frac{b_n}{10^n} = \frac8{89}

Answer and worked solution

Official answer: A, B, C

$\alpha,\beta=\frac{1\pm\sqrt5}2$ , so $a_n$ are the Fibonacci numbers (1, 1, 2, 3, 5, …).

(A) By induction, $\sum_{k\le n}a_k=a_{n+2}-1$ , since $a_{n+2}-1+a_{n+1}=a_{n+3}-1$ . True.

(B) The generating function gives
$\sum a_nx^n=\frac x{1-x-x^2};\qquad\text{at }x=\tfrac1{10}:\ \frac{0.1}{0.89}=\frac{10}{89}.$
True.

(C)
$a_{n-1}+a_{n+1}=\frac{\alpha^{n-1}(1+\alpha^2)-\beta^{n-1}(1+\beta^2)}{\alpha-\beta}=\alpha^n+\beta^n,$
using $1+\alpha^2=\alpha(\alpha-\beta)$ and $1+\beta^2=-\beta(\alpha-\beta)$ . Also $b_1=1=\alpha+\beta$ . True.

(D)
$\sum b_nx^n=\frac{x+2x^2}{1-x-x^2}=\frac{0.12}{0.89}=\frac{12}{89}\ne\frac8{89}.$
False.

(A), (B), (C)

Topics:Special series sums Roots and coefficients

← Previous question Next question →

JEE Advanced 2019 · Paper 1, Mathematics question 5