St. Ives Problem/Fibonacci Variant

Classic Problem

$7$ old women are travelling to Rome.

Each old woman has $7$ mules.

On each mule are $7$ sacks.

In each sack are $7$ loaves of bread.

In each loaf are $7$ knives.

Each knife has $7$ sheaths.

What is the total of all of these?

Thus we have:

Old women			$\ds 7 $
Mules	$\ds 7 \times 7 $	$\ds = $	$\ds 49 $
Sacks	$\ds 7 \times 7 \times 7 $	$\ds = $	$\ds 343 $
Loaves of bread	$\ds 7 \times 7 \times 7 \times 7 $	$\ds = $	$\ds 2401 $
Knives	$\ds 7 \times 7 \times 7 \times 7 \times 7 $	$\ds = $	$\ds 16 \, 807 $
Sheaths	$\ds 7 \times 7 \times 7 \times 7 \times 7 \times 7 $	$\ds = $	$\ds 117 \, 649 $

As with the St. Ives Problem, the total can be calculated using the Sum of Geometric Sequence:

$\ds $

$\ds 7^1 + 7^2 + 7^3 + 7^4 + 7^5 + 7^6$

$\ds $

$=$

$\ds 7 \times \paren {7^0 + 7^1 + 7^2 + 7^3 + 7^4 + 7^5}$

$\ds $

$=$

$\ds 7 \times \dfrac {7^6 - 1} {7 - 1}$

$\ds $

$=$

$\ds 137 \, 256$

$\blacksquare$

Old women			\(\ds 7 \)
Mules	\(\ds 7 \times 7 \)	\(\ds = \)	\(\ds 49 \)
Sacks	\(\ds 7 \times 7 \times 7 \)	\(\ds = \)	\(\ds 343 \)
Loaves of bread	\(\ds 7 \times 7 \times 7 \times 7 \)	\(\ds = \)	\(\ds 2401 \)
Knives	\(\ds 7 \times 7 \times 7 \times 7 \times 7 \)	\(\ds = \)	\(\ds 16 \, 807 \)
Sheaths	\(\ds 7 \times 7 \times 7 \times 7 \times 7 \times 7 \)	\(\ds = \)	\(\ds 117 \, 649 \)