Discussion

Explanation:

Let S = $\frac{1}{7}$ + $\frac{2}{7^2}$ + $\frac{1}{7^3}$ + $\frac{2}{7^4}$ + ... 

∴ S₁ = $\frac{1}{7}$ + $\frac{1}{7^3}$ + $\frac{1}{7^5}$ ... = $\frac{1/7}{1-1/7^2}$ = $\frac{7}{48}$

S₂ = $\frac{2}{7^2}$ + $\frac{2}{7^4}$ + $\frac{2}{7^6}$ + ...  = $\frac{2/7^2}{1-1/7^2}$ = $\frac{2}{48}$

Now, S = S₁ + S₂ = $\frac{7}{48}$ + $\frac{2}{48}$ = $\frac{9}{48}$

Hence, option (d).

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