Question

If we have \[\forall n\in N\] then, \[{{4}^{n}}-3n-1\] is divisible by
(A) 3
(B) 8
(C) 9
(D) 11

Answer 1

Hint: Modify the expression \[{{4}^{n}}-3n-1\] by writing 4 as the summation of 1 and 3. Now, expand \[{{\left( 1+3 \right)}^{n}}\] by using the binomial expansion formula, \[{{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+........{{+}^{n}}{{C}_{n}}{{x}^{n}}\] . We know that \[^{n}{{C}_{0}}=1\] and \[^{n}{{C}_{1}}=n\] . Now, solve it further and simplify it to get the number by which the expression \[{{4}^{n}}-3n-1\] is divisible.

Complete step-by-step solution
According to the question, we have \[\forall n\in N,{{4}^{n}}-3n-1\] and we have to find the number by which the expression \[{{4}^{n}}-3n-1\] is divisible, where n is any natural number.
The given expression = \[{{4}^{n}}-3n-1\] ………………………………..(1)
We have to simplify the above expression into a simpler form.
We can write 4 as the summation of 1 and 3, which is \[1+3=4\].
Now, on transforming the expression in equation (1), we get
\[={{\left( 1+3 \right)}^{n}}-3n-1\] ……………………………….(2)
We know the formula for the binomial expansion, \[{{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+........{{+}^{n}}{{C}_{n}}{{x}^{n}}\] …………………………………………………(3)
Now, on putting \[x=3\] in equation (3), we get
\[\Rightarrow {{\left( 1+3 \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}3{{+}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}\] …………………………………………….(4)
Using equation (4) and on substituting \[{{\left( 1+3 \right)}^{n}}\] by \[^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}3{{+}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}\] in equation (2), we get
\[{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}3{{+}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}-3n-1\] ……………………………………….(5)
We know that \[^{n}{{C}_{0}}=1\] and \[^{n}{{C}_{1}}=n\] ……………………………………….(6)
Now, using equation (6) and on replacing \[^{n}{{C}_{0}}\] and \[^{n}{{C}_{1}}\] by 1 and \[n\] in equation (5), we get
\[=1+n\times 3{{+}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}-3n-1\]
\[=1+3n{{+}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}-3n-1\] …………………………………………(7)
In the above equation, we can see that we have 1, \[3n\] , -1, and \[-3n\] which will cancel each other.
On cancelling 1 by -1 and \[3n\] by \[-3n\] in equation (7), we get
\[{{=}^{n}}{{C}_{2}}{{3}^{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n}}\] ……………………………………….(8)
Now, in equation (8), on taking the term \[{{3}^{2}}\] as common, we get
\[={{3}^{2}}\left( ^{n}{{C}_{2}}+........{{+}^{n}}{{C}_{n}}{{3}^{n-2}} \right)\] …………………………………………(9)
Clearly, we can see that the above equation is divisible by \[{{3}^{2}}\] that is 9.
Therefore, the expression \[{{4}^{n}}-3n-1\] for all \[n\in N\] is divisible by 9.
Hence, the correct option is (C).

Note: We can also solve this question by putting some random values of \[n\in N\] in the expression \[{{4}^{n}}-3n-1\] and then check by which number it gets completely divided. For instance, let us put \[n=2\] in the expression \[{{4}^{n}}-3n-1\] ,
\[\begin{align}
  & ={{4}^{2}}-3\times 2-1 \\
 & =9 \\
\end{align}\]
For \[n=2\] , the expression is divisible by 9.
Similarly, let us put \[n=3\] in the expression \[{{4}^{n}}-3n-1\] ,
\[\begin{align}
  & ={{4}^{3}}-3\times 3-1 \\
 & =54 \\
\end{align}\]
 For \[n=3\] , the expression is divisible by 9.
Therefore, the expression \[{{4}^{n}}-3n-1\] for all \[n\in N\] is divisible by 9.