If x ∈ 1,2,3 ,..., 9 and fn(x) =x x x ...x (n digits), then f2n (3)+fn(2) =

Question

If x ∈ 1,2,3 ,..., 9 and fn(x) =x x x ...x (n digits), then f2n (3)+fn(2) = (A) 2f2n(1) (B) f2n(1) (C) f2n (1) (D) -f2n(4)

Tardigrade Admin · Accepted Answer

fn(3) = 333 .... 3 (n digits) and fn(3) = 999 . . . 9 (n digits), f2n(2)= 222 ... 2 (n digits) ∴ f2n(3) +fn(2) = 12 . . . 2221 ((n + 1) digits) Answer cannot be (1), (3) or (4) fn(1) = 111. . . 1 (n digits) ∴ f2n(3)(1) = 122 . . . 21 ((n+ 1) digits).

Q. If $x \in {1, 2, 3, ..., 9}$ and $f_{n} (x) = x x x ... x$ ( $n$ digits), then $f_{n}^{2} (3) + f_{n} (2) =$

$2 f_{2 n} (1)$

$f_{n}^{2} (1)$

$f_{2 n} (1)$

$- f_{2 n} (4)$

$f_{n} (3) = 333....3$ (n digits) and $f_{n} (3) = 999...9$ (n digits),
$f_{n}^{2} (2) = 222...2$ (n digits)
$∴ f_{n}^{2} (3) + f_{n} (2) = 12...2221 ((n + 1)$ digits)
Answer cannot be (1), (3) or (4)
$f_{n} (1) = 111...1$ (n digits)
$∴ f_{n}^{2} (3) (1) = 122...21 ((n + 1)$ digits).

Q. If x∈{1,2,3,...,9} and fn​(x)=xxx...x (n digits), then fn2​(3)+fn​(2)=

2f2n​(1)

fn2​(1)

f2n​(1)

−f2n​(4)