If X = { 8^n - 7n - 1 : n in N } and Y = { 49 | vedclass

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(A) We have $8^n - 7n - 1 = (7 + 1)^n - 7n - 1$.Using the binomial expansion,$(7 + 1)^n = 1 + n(7) + \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + \bino

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$X \subseteq Y$

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(A) We have $8^n - 7n - 1 = (7 + 1)^n - 7n - 1$.Using the binomial expansion,$(7 + 1)^n = 1 + n(7) + \binom{n}{2}7^2 + \binom{n}{3}7^3 + \dots + \binom{n}{n}7^n$.So,$8^n - 7n - 1 = 1 + 7n + \binom{n}{2}49 + \binom{n}{3}7^3 + \dots + 7^n - 7n - 1$.$8^n - 7n - 1 = 49 \left[ \binom{n}{2} + \binom{n}{3}7 + \dots + 7^{n-2} ight]$.This shows that for $n \ge 2$,$8^n - 7n - 1$ is a multiple of $49$.For $n = 1$,$8^1 - 7(1) - 1 = 0$,which is $49 	imes 0$.For $n = 2$,$8^2 - 7(2) - 1 = 64 - 14 - 1 = 49$,which is $49 	imes 1$.Thus,every element of $X$ is a multiple of $49$,which is also an element of $Y$.Therefore,$X \subseteq Y$.

If $X = \{ 8^n - 7n - 1 : n \in N \}$ and $Y = \{ 49(n - 1) : n \in N \}$,then

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