Using binomial theorem, prove that

Question:

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

Solution:

$2^{3 n}-7 n-1=8^{n}-7 n-1$   ...(i)

Now,

$8^{n}=(1+7)^{n}$

$={ }^{n} C_{0}+{ }^{n} C_{1} \times 7^{1}+{ }^{n} C_{2} \times 7^{2}+{ }^{n} C_{3} \times 7^{3}+{ }^{n} C_{4} \times 7^{4}+\ldots+{ }^{n} C_{n} \times 7^{n}$

$\Rightarrow 8^{n}=1+7 n+49\left[{ }^{n} C_{2}+{ }^{n} C_{3} \times 7^{1}+{ }^{n} C_{4} \times 7^{2}+\ldots+{ }^{n} C_{n} \times 7^{n-2}\right]$

$\Rightarrow 8^{n}-1-7 n=49 \times($ An integer $)$

Now,

$8^{n}-1-7 n$ is divisible by 49

Or,

$2^{3 n}-1-7 n$ is divisible by 49 [From (1)]

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