Question

For \[n\in N\] , \[{{3}^{2n+2}}-{{2}^{3}}n-9\] is divisible by
(A) 3
(B) 9
(C) 64
(D) 81

Answer 1

Hint: First of all, transform the given expression \[\left( {{3}^{2n+2}}-{{2}^{3}}n-9 \right)\] as \[\left( {{9}^{n+1}}-8n-9 \right)\] . We can write 9 as the summation of 1 and 8, \[1+8=9\] . Now, replace 9 by \[\left( 1+8 \right)\] in the expression \[\left( {{9}^{n+1}}-8n-9 \right)\] and transform it as \[\left\{ {{\left( 1+8 \right)}^{n+1}}-8n-9 \right\}\] . We know the binomial expansion of \[{{\left( 1+x \right)}^{n}}\] , \[{{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+.........{{+}^{n}}{{C}_{n}}{{x}^{n}}\] . Now, replace x by 8 and \[n\] by \[\left( n+1 \right)\] in the expansion of \[{{\left( 1+x \right)}^{n}}\] . Now, assume \[\left( ^{n+1}{{C}_{2}}{{+}^{n+1}}{{C}_{3}}8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n-1}} \right)=k\] in the expansion of \[{{\left( 1+8 \right)}^{n+1}}\] . We know that \[^{n}{{C}_{0}}=1\] and \[^{n}{{C}_{1}}=n\] . Use this and get the value of \[^{n+1}{{C}_{0}}=1\] and \[^{n+1}{{C}_{1}}=n+1\] . Now, using this all, simply the expression \[\left\{ ^{n+1}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left( 64k \right) \right\}-8n-9\] . Solve it further and conclude the answer.

Complete step-by-step answer:
According to the question, we have the expression \[{{3}^{2n+2}}-{{2}^{3}}n-9\] where \[n\in N\] .
\[{{3}^{2n+2}}-{{2}^{3}}n-9\] …………………………(1)
We know the formula, \[{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}\] ………………………………..(2)
Using this formula, we can write, \[{{3}^{2n+2}}\] as \[{{\left( {{3}^{2}} \right)}^{\left( n+1 \right)}}\] . So,
\[{{3}^{2n+2}}={{\left( {{3}^{2}} \right)}^{n+1}}\] …………………………….(3)
Now, using equation (3) and replacing \[{{3}^{2n+2}}\] by \[{{\left( {{3}^{2}} \right)}^{n+1}}\] equation (1), we get
\[\begin{align}
  & ={{3}^{2n+2}}-{{2}^{3}}n-9 \\
 & ={{\left( {{3}^{2}} \right)}^{n+1}}-8n-9 \\
\end{align}\]
\[={{9}^{n+1}}-8n-9\] …………………………..(4)
We know that 9 can be written as the summation of 1 and 8, \[1+8=9\] ……………………………..(5)
Now, using equation (5) and replacing 9 by \[\left( 1+8 \right)\] in equation (4), we get
transforming equation (4), we get
\[={{\left( 1+8 \right)}^{n+1}}-8n-9\] …………………………………….(6)
We know the binomial expansion of \[{{\left( 1+x \right)}^{n}}\] , \[{{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}{{+}^{n}}{{C}_{3}}{{x}^{3}}+.........{{+}^{n}}{{C}_{n}}{{x}^{n}}\] ………………………………(7)
Now, replacing x by 8, \[n\] by \[\left( n+1 \right)\] in equation (7), we get
\[\begin{align}
  & \Rightarrow {{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8{{+}^{n+1}}{{C}_{2}}{{8}^{2}}{{+}^{n+1}}{{C}_{3}}{{8}^{3}}+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n+1}} \\
 & \Rightarrow {{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8{{+}^{n+1}}{{C}_{2}}\left( 64 \right){{+}^{n+1}}{{C}_{3}}\left( 64 \right)8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n+1}} \\
 & \Rightarrow {{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8{{+}^{n+1}}{{C}_{2}}\left( 64 \right){{+}^{n+1}}{{C}_{3}}\left( 64 \right)8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{2}}{{8}^{n-1}} \\
\end{align}\]
\[\Rightarrow {{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left\{ 64\left( ^{n+1}{{C}_{2}}{{+}^{n+1}}{{C}_{3}}8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n-1}} \right) \right\}\] …………………………..(8)
Let us assume that the value of \[\left( ^{n+1}{{C}_{2}}{{+}^{n+1}}{{C}_{3}}8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n-1}} \right)\] is \[k\] .
\[\left( ^{n+1}{{C}_{2}}{{+}^{n+1}}{{C}_{3}}8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n-1}} \right)=k\] …………………………………………(9)
Now, replacing \[\left( ^{n+1}{{C}_{2}}{{+}^{n+1}}{{C}_{3}}8+.........{{+}^{n+1}}{{C}_{n+1}}{{8}^{n-1}} \right)\] by \[k\] in equation (8), we get
\[\Rightarrow {{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left( 64k \right)\] …………………………………(10)
From equation (6), we have \[{{\left( 1+8 \right)}^{n+1}}-8n-9\] .
From equation (10), we have \[{{\left( 1+8 \right)}^{n+1}}{{=}^{n+1}}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left( 64k \right)\] .
Now, replacing \[{{\left( 1+8 \right)}^{n+1}}\] by \[\left\{ ^{n+1}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left( 64k \right) \right\}\] in equation (6), we get
\[={{\left( 1+8 \right)}^{n+1}}-8n-9\]
\[=\left\{ ^{n+1}{{C}_{0}}{{+}^{n+1}}{{C}_{1}}8+\left( 64k \right) \right\}-8n-9\] ……………………………………(11)
We know that \[^{n}{{C}_{0}}=1\] and \[^{n}{{C}_{1}}=n\] …………………..(12)
Now replacing n by \[\left( n+1 \right)\] in equation (12), we get
\[^{n+1}{{C}_{0}}=1\] and \[^{n+1}{{C}_{1}}=n+1\] ………………………………….(13)
Putting \[^{n+1}{{C}_{0}}=1\] and \[^{n+1}{{C}_{1}}=n+1\] in equation (11), we get
\[\begin{align}
  & =1+8\left( n+1 \right)+64k-8n-9 \\
 & =1+8n+8+64k-8n-9 \\
 & =9+8n+64k-8n-9 \\
 & =64k+8n-8n+9-9 \\
 & =64k \\
\end{align}\]
The value of \[{{3}^{2n+2}}-{{2}^{3}}n-9\] is equal to \[64k\] , \[{{3}^{2n+2}}-{{2}^{3}}n-9=64k\] .
Since 64 is a factor of \[64k\] so, we can say that \[64k\] is divisible by 64.
Therefore, the expression \[{{3}^{2n+2}}-{{2}^{3}}n-9\] is divisible by 64.
Hence, the correct option is (C).

Note: We can also solve this question by another method.
Since n is a natural number and a natural number starts from 1 so, n also takes value from 1.
Now, putting \[n=1\] in the expression \[{{3}^{2n+2}}-{{2}^{3}}n-9\] , we get
\[\begin{align}
  & ={{3}^{2\times 1+2}}-{{2}^{3}}\times 1-9 \\
 & ={{3}^{4}}-8-9 \\
 & =81-17 \\
 & =64 \\
\end{align}\]
This is divisible by 64.
Now, putting \[n=2\] in the expression \[{{3}^{2n+2}}-{{2}^{3}}n-9\] , we get
\[\begin{align}
  & ={{3}^{2\times 2+2}}-{{2}^{3}}\times 2-9 \\
 & ={{3}^{6}}-8\times 2-9 \\
 & =729-16-9 \\
 & =729-25 \\
 & =704 \\
\end{align}\]
This is also divisible by 64.
Therefore, for any value of \[n\in N\] , the expression \[{{3}^{2n+2}}-{{2}^{3}}n-9\] is divisible by 64.
Hence, the correct option is (C).