Question

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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{9^n=\left(1+8\right)^n}$

$\displaystyle\tt{\implies\,9^n=\,^nC_{0}(1)^n+\,^nC_{1}(1)^{n-1}\cdot8+\,^nC_{2}(1)^{n-2}\cdot8^{2}+\cdots+\,^nC_{n-1}(1)^{n-(n-1)}\cdot8^{n-1}+\,^nC_{n}8^{n}}$$\displaystyle\tt{\implies\,9^n=1+8n+\,^nC_{2}\,8^2+\cdots+\,^nC_{n-1}\,8^{n-1}+\,^nC_{n}\,8^{n}}$

$\displaystyle\tt{\implies\,9^n\cdot9=9+72n+9\left(\,^nC_{2}\,8^2+\cdots+\,^nC_{n-1}\,8^{n-1}+\,^nC_{n}\,8^{n}\right)}$

$\displaystyle\tt{\implies\,9^{n+1}=9+8n+64n+9\left(\,^nC_{2}\,8^2+\,^nC_{3}\,8^3+\cdots+\,^nC_{n-1}\,8^{n-1}+\,^nC_{n}\,8^{n}\right)}$

$\displaystyle\tt{\implies\,9^{n+1}-8n-9=64n+9\left(\,^nC_{2}\,8^2+\,^nC_{3}\,8^3+\cdots+\,^nC_{n-1}\,8^{n-1}+\,^nC_{n}\,8^{n}\right)}$

$\displaystyle\tt{\implies\,9^{n+1}-8n-9=64n+9\cdot8^2\left(\,^nC_{2}+\,^nC_{3}\,8+\cdots+\,^nC_{n-1}\,8^{n-3}+\,^nC_{n}\,8^{n-2}\right)}$

$\displaystyle\tt{\implies\,9^{n+1}-8n-9=64\left\{n+9\left(\,^nC_{2}+\,^nC_{3}\,8+\cdots+\,^nC_{n-1}\,8^{n-3}+\,^nC_{n}\,8^{n-2}\right)\right\}}$

The term inside the parenthesis is an integer, say m, so,

$\displaystyle\tt{\implies\,9^{n+1}-8n-9=64\left\{n+9m\right\}}$

Since n, m ∈ Z, so, n+9m ∈ Z,

Put n+9m=k,

$\displaystyle\tt{\implies\,9^{n+1}-8n-9=64\,k}$

Hence, $\tt{\blue{9^{n+1}-8b-9}}$ is divisible by 64

Answer 2

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