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Question
If \[X = \left\{ 8^n - 7n - 1: n \in N \right\} \text{ and } Y = \left\{ 49\left( n - 1 \right): n \in N \right\}\] \[X \subseteq Y .\]
Solution
Given:
\[X = \left\{ 8^n - 7n - 1: n \in N \right\} \text{ and } Y = \left\{ 49\left( n - 1 \right): n \in N \right\}\]
To prove:
\[X \subseteq Y\]
\[\text{ Let }: \]
\[ x_n = 8^n - 7n - 1, n \in N\]
\[ \Rightarrow x_1 = 8 - 7 - 1 = 0\]
\[\text{ For any n } \geqslant 2, \text{ we have }: \]
\[ x_n = 8^n - 7n - 1 = (1 + 7 )^n - 7n - 1\]
\[ \Rightarrow x_n = ^{n}{}{C}_0 + ^{n}{}{C}_1 . 7 + ^{n}{}{C}_2 . 7^2 + ^{n}{}{C}_3 . 7^3 + . . . +^{n}{}{C}_n . 7^n - 7n - 1\]
\[ \Rightarrow x_n = 1 + 7n + ^{n}{}{C}_2 . 7^2 + ^{n}{}{C}_3 . 7^3 + . . . + 7^n - 7n - 1 [ \because ^{n}{}{C}_0 = 1 and^{n}{}{C}_1 = n]\]
\[ \Rightarrow x_n = 7^2 {^{n}{}{C}_2 +^{n}{}{C}_3 . 7 + ^{n}{}{C}_4 7^2 + . . . + ^{n}{}{C}_n . 7^{n - 2} }\]
\[ \Rightarrow x_n = 49{^{n}{}{C}_2 + ^{n}{}{C}_3 . 7 + ^{n}{}{C}_4 7^2 + . . . + ^{n}{}{C}_n . 7^{n - 2} }\]
\[\text{ Thus, x_n is some positive integral multiple of 49 for all } n \geqslant 2 . \]
\[X \text{ consists of all those positive integral multiples of 49 that are of the form } 49{ ^{n}{}{C}_2 +^{n}{}{C}_3 . 7 + ^{n}{}{C}_4 7^2 + . . . +^{n}{}{C}_n . 7^{n - 2} } \text{ along with zero } . \]
\[Y = {49(n - 1): n \text{ in } N} \text{ implies that it consists of all integral multiples of 49 along with zero } . \]
\[ \therefore X \subseteq Y\]
\[\]
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