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

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Solution

$\begin{matrix} 2^{3 n} - 7 n - 1 8^{n} - 7 n - 1 = {(1 + 7)}^{n} - 7 n - 1 U s e b i n o m i a l t h e o r e m {(1 + x)}^{n} = 1 +^{n} c_{1} . x +^{n} c_{2} x^{2} + . . . . . . . . + x^{n} {(1 + 7)}^{n} = 1 +^{n} c_{1} .7 +^{n} c_{2} {.7}^{2} + . . . . . . . + 7^{n} T h e n {(1 + 7)}^{n} - 7 n - 1 = 1 +^{n} c_{1} .7 +^{n} c_{2} {.7}^{2} + . . . . + 7^{n} - 7 n - 1 =^{n} c_{2} {.7}^{2} +^{n} c_{3} {.7}^{3} + . . . . . . . + 7^{n} (^{n} c_{1} = n) = 7^{2} (^{n} c_{2} +^{n} c_{3} .7 + . . . . + 7^{n - 2}) = 49 \times int e g e r = 2^{3 n} - 7 n - 1 i s d i v i s i b l e b y 49 a l w a y s . \end{matrix}$

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