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

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Question

(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} -

Vedclass · Accepted Answer

$X \subseteq Y$

Vedclass · Answer

(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$

${ = ^n}{C_2}{7^2}{ + ^n}{C_3}{7^3} + ..{ + ^n}{C_n}{7^n}$,${(^n}{C_0}{ = ^n}{C_n},{\,^n}{C_1}{ = ^n}{C_{n - 1}}\,{m{etc}}{m{.)}}$

$ = 49{[^n}{C_2}{ + ^n}{C_3}(7) + ......{ + ^n}{C_n}{7^{n - 2}}]$

$	herefore$ ${8^n} - 7n - 1$ is a multiple of $49$ for $n \ge 2$

For $n = 1$, ${8^n} - 7n - 1 = 8 - 7 - 1 = 0$;

For $n = 2,$ ${8^n} - 7n - 1 = 64 - 14 - 1 = 49$

$	herefore$ ${8^n} - 7n - 1$ is a multiple of $49$ for all $n \in N.$

$	herefore $ $X$ contains elements which are multiples of $49$ and clearly $Y$ contains all multiplies 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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