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જો $X = \{ {8^n} - 7n - 1:n \in N\} $ અને $Y = \{ 49(n - 1):n \in N\} ,$ તો . . ..
$X \subseteq Y$
$Y \subseteq X$
$X = Y$
એકપણ નહીં.
Solution
(a) Since ${8^n} – 7n – 1 = {(7 + 1)^n} – 7n – 1$
$ = {7^n}{ + ^n}{C_1}{7^{n – 1}}{ + ^n}{C_2}{7^{n – 2}} + …..{ + ^n}{C_{n – 1}}7{ + ^n}{C_n} – 7n – 1$
${ = ^n}{C_2}{7^2}{ + ^n}{C_3}{7^3} + ..{ + ^n}{C_n}{7^n}$,${(^n}{C_0}{ = ^n}{C_n},{\,^n}{C_1}{ = ^n}{C_{n – 1}}\,{\rm{etc}}{\rm{.)}}$
$ = 49{[^n}{C_2}{ + ^n}{C_3}(7) + ……{ + ^n}{C_n}{7^{n – 2}}]$
$\therefore$ ${8^n} – 7n – 1$ is a multiple of $49$ for $n \ge 2$
For $n = 1$, ${8^n} – 7n – 1 = 8 – 7 – 1 = 0$;
For $n = 2,$ ${8^n} – 7n – 1 = 64 – 14 – 1 = 49$
$\therefore$ ${8^n} – 7n – 1$ is a multiple of $49$ for all $n \in N.$
$\therefore $ $X$ contains elements which are multiples of $49$ and clearly $Y$ contains all multiplies of $49$. $X \subseteq Y$.
