<s>[INST] Comparing sets $\{8^n-7n-1: n \in \Bbb{N}\}$ and $\{49(n-1): n \in \Bbb{N}\}$ 
If
$$\begin{align}
X&=\{8^n-7n-1: n \in \Bbb{N}\} \\
Y&=\{49(n-1): n \in \Bbb{N}\}
\end{align}$$
then, 
a) $X\subset Y \qquad$ b) $Y\subset X\qquad$ c) $X=Y\qquad$d) none of these

I know this question can be solved by taking $X=8^n-7n-1$ and splitting $8^n$ into $(7+1)^n$ and then apply binomial theorem as follows:
Given,
$$\begin{align}X &=8^n−7n−1 \\[4pt]
&=(1+7)^n−7n−1 \\[4pt]
&=1+7n+\frac{n(n-1)}{2}+\cdots+7^n-7n-1 \\[4pt]
&=\frac{n(n-1)}{2}7^2+\cdots+7^n \\[4pt]
&=49\left[\frac{n(n-1)}{2}+\cdots+7^{n-2}\right]
\end{align}$$
Hence, the set $X$ will be some specific multiples of $49$.

$Y=49(n-1)$. Hence, the set $Y$ will be all multiples of $49$. So, it will contain the elements of $X$ too.
So, $$X\subset Y$$

But is there any alternate/simple method to solve this question without using binomial theorem?

 [/INST] $8^n - 7n - 1 = (8^n - 1) - 7n = (8-1)(8^{n-1} + ..... + 8 + 1) - 7n = 7[(8^{n-1} + .... + 8 + 1) - n]$.
So $7|8^n - 7n - 1$.
And $8^n \equiv 1^n \equiv 1 \pmod 7$ so $(8^{n-1} + .... + 8 + 1) - n\equiv (1+1+1+....+1) -n \equiv n - n \equiv 8 \pmod 7$
So $7|(8^{n-1} + .... + 8 + 1) - n$ and $49|8^n - 7n - 1=7[(8^{n-1} + .... + 8 + 1) - n]$.
As $Y$ is the set of all non negative multiple of $49$ and $8^n-7n - 1$ is a non-negative multiple of $49$ we know $X\subset Y$.
But as $n$ gets large $8^n -7n -1$ will increase by larger amounts than $49$ and not all multiples of $49$ can possibly be covered by $8^n-7n-1$.
Indeed for $n = 2$ we have $8^2 - 7\cdot 2 -1 = 49$ but for $n =3$ we have $8^2 -3\cdot 7 -1 = 490$ and we never have any multiples of $49$ between $49=49\cdot 1$ and $490 = 49\cdot 10$.
And furthermore  If $A_k = 8^k - 7k - 1$ then $D=A_{k+1} - A_k = 7\cdot 8^{k+1} - 7$. and $k> 1$ then $D > 49$ and $A_n$ is increasing and by more than $49$.
So $Y\not \subset X$.
</s>