Question

If X=\left\\{ {{8}^{n}}-7n-1:n\in N \right\\} and Y=\left\\{ 49\left( n-1 \right):n\in N \right\\}, then
(a) $X\subseteq Y$
(b) $Y\subset X$
(c) $X=Y$
(d) None of these

Answer 1

Hint: First of all we will find the simple form of X using Binomial theorem i.e. ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}+{}^{n}{{C}_{1}}\left( x \right)+{}^{n}{{C}_{2}}{{\left( x \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( x \right)}^{n}}$and then we will compare values of X and Y, if values of X will be present in Y we will write $X\subseteq Y$, if values of Y will be in X then we will write $Y\subset X$and if all the values of X and Y will be same then we will write $X=Y$and if nothing matches then we will go for None of these, so, by finding some values of X and Y we will solve the answer.

Complete step-by-step answer:
In the question we are given that X=\left\\{ {{8}^{n}}-7n-1:n\in N \right\\} and Y=\left\\{ 49\left( n-1 \right):n\in N \right\\}and we are asked to find the relation between X and Y. So, first of all X and Y can be shown as,
X=\left\\{ {{8}^{n}}-7n-1 \right\\} ………………………(i)
Y=\left\\{ 49\left( n-1 \right) \right\\} ………………………..(ii)
Now, we will find the simplified form of expression (i) so the equation can also be expressed as,
X=\left\\{ {{\left( 1+7 \right)}^{n}}-7n-1 \right\\} ……………(iii)
Now, the Binomial form of equation ${{\left( 1+x \right)}^{n}}$ can be given as,
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}+{}^{n}{{C}_{1}}\left( x \right)+{}^{n}{{C}_{2}}{{\left( x \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( x \right)}^{n}}$ ………………(iv)
So, we can also write ${{\left( 1+7 \right)}^{n}}$ in the form of expression (iii) which can be given as,
${{\left( 1+7 \right)}^{n}}={}^{n}{{C}_{0}}{{\left( 1 \right)}^{n}}+{}^{n}{{C}_{1}}\left( 7 \right)+{}^{n}{{C}_{2}}{{\left( 7 \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n}}-7n-1$
Now, we know that, ${}^{n}{{C}_{1}}=\dfrac{n!}{1!\left( n-1 \right)!}=n$, so using this formula we can write, ${}^{n}{{C}_{1}}\left( 7 \right)$ as $7n$ and ${}^{n}{{C}_{0}}=1$, so substituting these values we will get,
$\Rightarrow {{\left( 1+7 \right)}^{n}}=1+7n+{}^{n}{{C}_{2}}{{\left( 7 \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n}}$
Now, substituting this value in expression (iii) we will get,
X=\left\\{ 1+7n+{}^{n}{{C}_{2}}{{\left( 7 \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n}}-7n-1 \right\\}
$\Rightarrow X=\left( {}^{n}{{C}_{2}}{{\left( 7 \right)}^{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n}} \right)$
Now, on taking ${{7}^{2}}$ common from the whole equation we will get,
$\Rightarrow X={{\left( 7 \right)}^{2}}\left[ {}^{n}{{C}_{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n-2}} \right]$
$\Rightarrow X=49\left[ {}^{n}{{C}_{2}}+.....+{}^{n}{{C}_{n}}{{\left( 7 \right)}^{n-2}} \right]$…….(v)
From expression (v) we can say that X is divisible by 49.
Now considering expression (ii) i.e. Y=\left\\{ 49\left( n-1 \right):n\in N \right\\}, fom expression (ii) it can be said that all values of Y will be divisible by 49.
So, from this it can be said that X is divisible by 49 and Y has all multiples of 49 as well as all values of X will be included in Y, so we can say that $X\subseteq Y$.
Hence, option (a) is correct.

Note: An alternative method to solve the problem is by considering different values of n i.e. $n=1,2,3......$ and substituting them in equations X=\left\\{ {{8}^{n}}-7n-1:n\in N \right\\} and Y=\left\\{ 49\left( n-1 \right):n\in N \right\\}as,
For $n=1$, $X=0$and $Y=0$
For $n=2$,$X=64-14-1=49$ and $Y=49\left( 2-1 \right)=49$
For $n=3$, $X=512-21-1=490$ and $Y=49\left( 3-1 \right)=98$
The values of X can als be written as, X=\left\\{ 0,49,490...... \right\\} and values of Y can be given as, Y=\left\\{ 0,49,98...... \right\\}.
Now, the tenth value of n will be 490 in Y, so, from this we can say that all values of X are present in Y, therefore from this we can say that $X\subseteq Y$. But this method consumes more time as we have to find values for each value of n, so, the method of Binomial theorem shown in solution is more preferable.