Question
If $\text{X}=\{8^\text{n}-7\text{x}-1:\text{n}\in\text{N}\}$ and $\text{Y}=\{49(\text{n}-1):\text{n}\in\text{N}\},$ then prove thet $\text{X}\subseteq\text{Y}.$

Answer

$\text{X}=\{8^\text{n}-7\text{x}-1:\text{n}\in\text{N}\}$ $\text{Y}=\{49(\text{n}-1):\text{n}\in\text{N}\}$ In order to show that $\text{X}\subseteq\text{Y}$ we shoe the every element of X is an element of Y. So let $\text{x}\in\text{X}\Rightarrow\text{x}=8^\text{n}-7\text{m}-1$ for same $\text{m}\in\text{N}$ $\Rightarrow\text{x = (1 + 7)}^\text{m}- 7\text{m} - 1$ $=(^\text{m}\text{C}_01^\text{m}+^\text{m}\text{C}_11^\text{m-1}7+...+^\text{m}\text{C}_\text{m-1}1^17^\text{m-1}+^\text{m}\text{C}_\text{m}7^\text{m})-7\text{m}-1$ [using binomail expansion] $=1+7\text{m}+^\text{m}\text{C}_27^2+^\text{m}\text{C}_37^3+...+^\text{m}\text{C}_\text{m}7^\text{m}-7\text{m}-1$ $=\ ^\text{m}\text{C}_27^2+\ ^\text{m}\text{C}_37^3+...+\ ^\text{m}\text{C}_\text{m}7^\text{m}$ $=49(^\text{m}\text{C}_2+^\text{m}\text{C}_3+...+^\text{m}\text{C}_\text{m}7^\text{m}),\ \text{m}\geq2 $ $=49\text{t}_\text{m},\ \text{m}\geq2,$ where $\text{t}_\text{m}=^\text{m}\text{C}_2+^\text{m}\text{C}_37+...+^\text{m}\text{C}_\text{m}7^\text{m-2}$ Is some positive integer depending on $\text{m}\geq2$ For $\text{m} = 1$$\text{x} = 1^8 - 7 × 1 - 1$
$=8 - 8$
$= 0$
Hence, X contains all positive integral multiples of 49. Also, Y consistes all positive integral multiples of 49, including 0, for n = 1. Thuse, we coclude that $\text{X}\subseteq\text{Y}.$

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