First slide

Introduction to sets

Question

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

Easy

A

$X \subseteq Y$
B

$Y \subseteq X$
C

$X = Y$
D

None of these

Solution

Since $8^{n} - 7 n - 1 = {(7 + 1)}^{n} - 7 n - 1$
$= 7^{n} + n_{C_{1}} 7^{n - 1} + n_{C_{2}} 7^{n - 2} + . . . + n_{C_{n - 1}} 7 + n_{C_{n}} - 7 n - 1$
$= n_{C_{2}} 7^{2} + n_{C_{3}} 7^{3} + . . + n_{C_{n}} 7^{n}, (n_{C_{0}}, n_{C_{1}} = n_{C_{n - 1}} etc .)$
$= 49 [n_{C_{2}} + n_{C_{3}} (7) + . . . + n_{C_{n}} 7^{n - 2}]$
$∴ 8^{n} - 7 n - 1$ is a multiple of 49 for $n \geq 2$
For $n = 1, 8^{n} - 7 n - 1 = 8 - 7 - 1 = 0;$
For $n = 2, 8^{n} - 7 n - 1 = 64 - 14 - 1 = 49$
$∴ 8^{n} - 7 n - 1$ is a multiple of 49 for all $n \in N .$
$∴ X$ contains elements which are multiples of 49 and clearly Y contains all multiplies of $49 ∴ X \subseteq Y .$

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