Using binomial theorem, prove that
Question:

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

Solution:

To prove: $\left(2^{3 n}-7 n-1\right)$ is divisible by 49, where $n N$

Formula used: $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

$\left(2^{3 n}-7 n-1\right)=\left(2^{3}\right)^{n}-7 n-1$

$\Rightarrow 8^{n}-7 n-1$

$\Rightarrow(1+7)^{n}-7 n-1$

$\Rightarrow{ }^{n} C_{0} 1^{n}+{ }^{n} C_{1} 1^{n-1} 7+{ }^{n} C_{2} 1^{n-2} 7^{2}+\ldots \ldots+{ }^{n} C_{n-1} 7^{n-1}+{ }^{n} C_{n} 7^{n}-7 n-1$

$\Rightarrow{ }^{n} C_{0}+{ }^{n} C_{1} 7+{ }^{n} C_{2} 7^{2}+\ldots \ldots+{ }^{n} C_{n-1} 7^{n-1}+{ }^{n} C_{n} 7^{n}-7 n-1$

$\Rightarrow 1+7 n+7^{2}\left[{ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+{ }^{n} C_{n-1} 7^{n-3}+{ }^{n} C_{n} 7^{n-2}\right]-7 n-1$

$\Rightarrow 7^{2}\left[{ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+{ }^{n} C_{n-1} 7^{n-3}+{ }^{n} C_{n} 7^{n-2}\right]$

$\Rightarrow 49\left[{ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+{ }^{n} C_{n-1} 7^{n-3}+{ }^{n} C_{n} 7^{n-2}\right]$

$\Rightarrow 49 K$, where $K=\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+{ }^{n} C_{n-1} 7^{n-3}+{ }^{n} C_{n} 7^{n-2}\right)$

Now, $\left(2^{3 n}-7 n-1\right)=49 K$

Therefore $\left(2^{3 n}-7 n-1\right)$ is divisible by 49

 

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