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

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 ${{4}^{n}}-3n-1$ ,

& ={{4}^{2}}-3\times 2-1 \\\ & =9 \\\ \end{aligned}$$ For $$n=2$$ , the expression is divisible by 9. Similarly, let us put $$n=3$$ in the expression $${{4}^{n}}-3n-1$$ , $$\begin{aligned} & ={{4}^{3}}-3\times 3-1 \\\ & =54 \\\ \end{aligned}$$ For $$n=3$$ , the expression is divisible by 9. Therefore, the expression $${{4}^{n}}-3n-1$$ for all $$n\in N$$ is divisible by 9.