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

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(A) Given $X = \{ 8^n - 7n - 1 : n \in N \}$ and $Y = \{ 49(n - 1) : n \in N \}$.Using the binomial expansion,$8^n = (1 + 7)^n = 1 + ^nC_1(7) + ^nC_2(

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$X \subseteq Y$

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(A) Given $X = \{ 8^n - 7n - 1 : n \in N \}$ and $Y = \{ 49(n - 1) : n \in N \}$.Using the binomial expansion,$8^n = (1 + 7)^n = 1 + ^nC_1(7) + ^nC_2(7^2) + ^nC_3(7^3) + \dots + ^nC_n(7^n)$.Therefore,$8^n - 7n - 1 = (1 + 7n + ^nC_2(7^2) + ^nC_3(7^3) + \dots + 7^n) - 7n - 1 = ^nC_2(7^2) + ^nC_3(7^3) + \dots + 7^n$.This simplifies to $49 	imes [^nC_2 + ^nC_3(7) + \dots + 7^{n-2}]$.For $n=1$,$8^1 - 7(1) - 1 = 0$.For $n=2$,$8^2 - 7(2) - 1 = 64 - 14 - 1 = 49$.For $n=3$,$8^3 - 7(3) - 1 = 512 - 21 - 1 = 490 = 49 	imes 10$.Thus,every element in $X$ is a multiple of $49$ (including $0$).$Y = \{ 0, 49, 98, 147, \dots \}$ represents all non-negative multiples of $49$.Since every element of $X$ is a multiple of $49$,$X \subseteq Y$.

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

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