Question

If n is a positive integer, prove that $3^{3\text{n}}-26\text{n}-1$ is divisible by 676.

Answer

$3^{3\text{n}}-26\text{n}-1$

$=(3^3)^\text{n}-26\text{n}-1$

$=27^\text{n}-26\text{n}-1$

$=(1+26)^\text{n}-26\text{n}-1$

$\Big({^\text{n}\text{C}}_0+{^\text{n}\text{C}}_1(26)^1+{^\text{n}\text{C}}_2(26)^2+.....+676(26)^{\text{n}-2}\Big)-26\text{n}-1$

$=676\Big({^\text{n}\text{C}}_2+......+(26)^{\text{n}-2}\Big)$

$\therefore3^{3\text{n}}-26\text{n}-1$ is divisible for $\text{n}\in\text{N}.$

Hence, proved

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