Q43 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

if $3^{2 n+2}-8 n-9$ is divisible by $2^p \forall n \in N$, then the maximum value of $p$ is

  • A
    8
  • B
    7
  • Correct answer
    C
    6
  • D
    9

Solution

$3^{2 n+2}-8 n-9=9(1+8)^n-8 n-9$ $\begin{aligned} & =9\left[1+{ }^n \mathrm{C}_1 8+{ }^n \mathrm{C}_2 8^2+{ }^n \mathrm{C}_3 8^3+\ldots+{ }^n \mathrm{C}_n 8\right]-8 n-1 \\ & =72 n-8 n+9\left[{ }^n \mathrm{C}_2 8^2+{ }^n \mathrm{C}_3 8^3+\ldots+{ }^n \mathrm{C}_n 8^n\right] \\ & =64 n+9 \times 64\left[{ }^n \mathrm{C}_2+{ }^n \mathrm{C}_3 8+\ldots+8^{n-2}\right]\end{aligned}$ $=54 \times k$ (where $k$ is the integer) Hence $\left(3^{n+2}-8 n-9\right)$ is divisible by $64=2^6$ $\therefore p=6$

Hence, the correct answer is option: 'C'.

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