By using Binomial theorem, prove that:
$3^{3 n} - 26 n - 1$ is divisible by $676$ , for all n $\in$ N.

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Solution

$3^{3 x} - 26 n - 1 = (3^{3})^{n} - 26 n - 1 = 27^{n} - 26 n - 1 = (1 + 26)^{n} - 26 n - 1 = [1 + n \cdot 26 +^{n} C_{2} 26^{2} + (h i g h e r t e r m) + . . . . .] - 26 n - 1 = 26^{2} [^{n} C_{2} +^{n} C_{3} 26 + . . . . . . h i g h e r t e r m s] c l e a r l y i t w i l l b e d i v i s i b l e b y 26^{2} o r 676 f o r a l l n \in N$

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