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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?

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    Hint $-$ using difference of powers: $;8^n-1-7n = (8-1)\left(\underbrace{(8^{n-1}+8^{n-2}+\dots+1)}_{\equiv, 1 + 1 + \dots +1 ,=, n \pmod{7}}-n\right),$. – dxiv Jul 17 '21 at 05:25
  • It's immediate that $8^n-7n-1=0\pmod 7$. Then you could "brute-force" the rest by verifying $8^n-7n-1=0\pmod {49}$ by hand for $n\in{0,\ldots,6}$, which implies it holds for all $n$. – Milten Jul 17 '21 at 05:31
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    @Milten yes it is given in the question that n is a natural no i.e{1,2,3,4........}. I have edited the question –  Jul 17 '21 at 06:50
  • Yes, it's the Illustration 6 of chapter 5 Sets of RD Sharma. In that book it has been solved using binomial theorem which is a bit confusing for me, so I wanted to know a alternate method for this. –  Jul 17 '21 at 18:35

4 Answers4

1

Here's an alternative method, using dxiv's suggestion to invoke the difference-of-powers identity:

Let $n\in\mathbb N.$ $$8^n-7n-1\\=(8-1)(8^0+8^1+\ldots+8^{n-1})-7n$$$$=7(8^0+8^1+\ldots+8^{n-1}-n).\tag1$$ Working $\mod7,$ $$8^0+8^1+\ldots+8^{n-1}-n\\\equiv n\times1-n$$$$=0.\tag2$$ From $(1)$ and $(2),$ and since $8^n-7n-1$ is increasing with $n,$ and equals $0, 49$ and $490$ when $n$ equals $1,2$ and $3,$ respectively, $8^n-7n-1$ is some nonnegative multiple of $49$ but is not $2\times49.$

Hence, for some proper subset $A$ of $\mathbb N,$ $$X=\{8^n-7n-1\mid n \in \mathbb N\}\\=\{49(k-1)\mid k \in A\}.$$ But $$\{49(k-1)\mid k \in A\}\\\subset\{49(n-1)\mid n \in \mathbb N\}=Y.$$ Therefore $$X\subset Y.$$

ryang
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Induction:

$n=0$: $49|8^0-7\cdot 0-1$

The difference between $n$ and $n+1$ is $8^{n+1}-8^n-7=7(8^n-1)$ which is divisible by $49$, so if $49$ divides the $n$th term, it divides the following one.

Eric
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Putting $n=1,2,3,4,5,…$, we obtain: $X=\{0,49,490,4067,32732,…\}$.

Similarly, putting $n=1,2,3,…,11,…,84,…,669$, we have: $Y=\{0,49,98,...,490,…,4067,…,32732,…\}$.

$X=\left\{8^{n}-7 n-1 \mid n \in N\right\}=\{0,49,490, \ldots\}$

$Y=\{49 n-49 \mid n \in N\}=\{0,49,98,147, \ldots, 490, \ldots\}$

Clearly, every element of $X$ is in $Y$ but every element of $Y$ is not in $X$.

$\therefore \quad X \subset Y$

0

$8^n - 7n - 1 = (8^n - 1) - 7n = (8-1)(8^{n-1} + ..... + 8 + 1) - 7n = 7[(8^{n-1} + .... + 8 + 1) - n]$.

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 $Y\not \subset X$.

fleablood
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