Question
Using binomial theorem, prove that $\left(2^{3 n }-7 n -1\right)$ is divisible by 49 , where $n \in N$

Answer

To prove: $\left(2^{3 n}-7 n-1\right)$ is divisible by 49 , where $n \in N$
$\begin{array}{l}\left(2^{3 n}-7 n-1\right)=\left(2^3\right) n-7 n-1 \\ =8^n-7 n-1 \\ =(1+7)^n-7 n-1\end{array}$
Now using binomial theorem.. 
$\begin{array}{l}\Rightarrow{ }^n C_0 1^n+{ }^n C_1 1^{n-1} 7+{ }^n C_2 1^{n-2} 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1 \\ ={ }^n C_0+{ }^n C_1 7+{ }^n C_2 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1 \\ =1+7 n+7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right]-7 n-1 \\ =7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right] \\ =49\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right] \\ =49 K, \text { where } K=\left({ }^n C_2+{ }^n C_3 7+\ldots++{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right)\end{array}$
Now, $\left(2^{3 n}-7 n-1\right)=49 K$
Therefore $\left(2^{3 n}-7 n-1\right)$ is divisible by 49 .

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free