MCQ
For every integer $\text{n}\geq1,$ $(3^{2n}-1)$ is always divisible by:
- A$2^{n2}$
- B$2^{n+4}$
- ✓$2^{n+2}$
- D$2^{n+3}$
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$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2}\right)+$
$\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2.3^2}-\frac{1}{3^3}\right)+$
$\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\ldots$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to....