1. ## sets

if x={8^n - 7n - 1} and y={49(n-1)} prove that x is a subset of y

2. Originally Posted by grgrsanjay
if x={8^n - 7n - 1} and y={49(n-1)} prove that x is a subset of y
So presumably n is a positive integer.

You need to show that $8^n - 7n - 1 \equiv 0\ (\text{mod }49)$

(The expression also must be non-negative but this is trivial.)

Use mathematical induction. Base case, n = 1.

$8^1 - 7(1) - 1 = 0 \equiv 0\ (\text{mod }49)$

Induction step.

$8^{n+1} - 7(n+1) - 1 \equiv 8 \cdot 8^n - 7n - 7 - 1 \equiv (8)(7n+1) - 7n - 8 \equiv$
$56n + 8 - 7n - 8 \equiv 49n \equiv 0\ (\text{mod }49)$

3. Hello, grgrsanjay!

If $x\:=\:8^n - 7n - 1\,\text{ and }\,y\:=\:49(n-1)$, prove that: . $x \subset y.$

We want to show that $x \:=\:8^n - 7n - 1$ is always a multiple of 49.

We have: . $x\;=\;8^n - 7n - 1$

. . . . . . . . . $=\;(7+1)^n - 7n - 1$

. . . . . . . . . $=\;\bigg[7^n + {n\choose n-1}7^{n-1} + {n\choose n-2}7^{n-2} + \hdots + {n\choose2}7^2 + {n\choose1}7 + 1
\bigg] - 7n - 1$

. . . . . . . . . $=\;7^n + {n\choose n-1}7^{n-1} + {n\choose n-2}7^{n-2} + \hdots + {n\choose2}7^2$

. . . . . . . . . $=\;7^2\underbrace{\bigg[7^{n-2} + {n\choose n-1}7^{n-3} + {n\choose n-2}7^{n-4} + \hdots + {n\choose2}\bigg]}_{\text{This is an integer}}$

Therefore: . $x$ is a multiple of 49.