For all positive integers n- and n - 1-let a n -1- a n a n -12 and a 0-1- a 1-8 then a 10- -A- 2210-1B- 222-1C- 228-1D- 2211-1

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For all posit...

Question

For all positive integers n, and $n \geq 1$ ,
let $a_{n + 1} = a_{n} a_{n - 1}^{2}$ and $a_{0} = 1, a_{1} = 8$ then $a_{10} = ?$

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Solution

The correct option is B

Method 1: Conventional Approach

Given : $a_{n + 1} = a_{n} a_{n - 1}^{2}$
$a_{n + 1} a_{n} = (a_{n} a_{n - 1})^{2^{1}}$
$(a_{n} a_{n - 1})^{2} - (a_{n - 1} a_{n - 2})^{2^{2}}$
$. .$
$. .$
$(a_{2} a_{1})^{2^{n - 1}} = (a_{1} a_{0})^{2^{n}}$
Multiplying all equations, we get
$a_{n + 1} a_{n} = (a_{1} a_{0})^{2^{n}} = 8^{2^{n}}$
Now $a_{10} a_{9} = 8^{2^{9}}, a_{9} a_{8} = 8^{2^{8}} . . . . . .$
$a_{10} = 8^{2^{9} - 2^{8} + 2^{7} - 2^{6} + 2^{5} - 2^{4} + 2^{3} - 2^{2}} \times a_{2}$ $(a_{2} = {8.1}^{2} = 8)$
$⟹ a_{10} = 8^{⎛ ⎝ \frac{2^{10} - 1}{3} ⎞ ⎠}$
$⟹ a_{10} = 2^{(2^{10} - 1)}$

Method 2: Shortcut: Double Substitution

This technique involves writing the answer choice in terms of variables. We need to find $a_{10}$ .

Thus n=10. Now rewrite all the options in terms of n.

$a . 2^{(2^{n + 1} - 1)} b . 2^{(2^{n} - 1)} c . 2^{(2^{n - 1} - 1)} d . 2^{(2^{n - 2} - 1)}$
At n=2 in the question, $a_{2} = a_{1} \times a_{0}^{2} = 8 \times 1 = 8$
Now put n=2 in each of the answer choices and check where you are getting 8.
$a . 2^{(2^{3} - 1)} \neq 8 b . 2^{(2^{2} - 1)} = 8 c . 2^{2} \neq 8 d . 2^{0} \neq 8$
Thus, options a, c, d can be eliminated. Answer is option (b).

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For all positive integers n, and n≥1, let an+1=ana2n−1 and a0=1, a1=8 then a10=?

For all positive integers n, and $n \geq 1$ ,
let $a_{n + 1} = a_{n} a_{n - 1}^{2}$ and $a_{0} = 1, a_{1} = 8$ then $a_{10} = ?$