CAT 2024 Slot 2 · Quant · Question 8

The general term is {\left( \frac{1}{5}\right)}^{n}\left({\left( \frac{1}{5}\right)}^{n}-{\left( \frac{1}{7}\right)}^{n}\right).

Expanding the bracket splits it into {\left( \frac{1}{25}\right)}^{n}-{\left( \frac{1}{35}\right)}^{n}.

So the series is the difference of two infinite geometric series, one with ratio \frac{1}{25} and the other with ratio \frac{1}{35}, each starting at n=1.

An infinite geometric series with first term r and ratio r sums to \frac{r}{1-r}.

The first sums to \frac{\frac{1}{25}}{1-\frac{1}{25}}=\frac{1}{24}.

The second sums to \frac{\frac{1}{35}}{1-\frac{1}{35}}=\frac{1}{34}.

The required value is \frac{1}{24}-\frac{1}{34}=\frac{34-24}{816}=\frac{10}{816}.

Reducing, \frac{10}{816}=\frac{5}{408}, which is the first option.

Shortcut. The factor \frac{1}{k} in front of a bracket with \frac{1}{k} inside is a hint to multiply through first. Once the term reads \frac{1}{{25}^{n}}-\frac{1}{{35}^{n}}, the standard result \frac{1}{k-1} for ∑\frac{1}{{k}^{n}} gives \frac{1}{24}-\frac{1}{34} on sight.