Quick GP sum

Summing a finite geometric progression (GP) is a core skill in the HSC Sequences topic. This problem lets you practise identifying the first term, common ratio, and number of terms, then applying the finite sum formula neatly. The same technique generalises to any finite GP you'll meet in exams. Recognising the pattern quickly ensures you can handle any finite GP on your exam paper.

Problem Statement

Find

$$1+\frac12+\frac14+\frac18+\cdots+\frac1{128}.$$

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Hints

This is a finite GP with $a=1$, $r=\frac12$, and $8$ terms.

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Solutions

We recognise the series as a finite geometric progression. The first term is $a=1$, and the common ratio is $r=\frac12$ because each term is half the previous one. The last term is $\frac1{128} = (\frac12)^7$, so the series contains $n=8$ terms (the first term corresponds to $r^0$). The sum of the first $n$ terms of a GP is given by $S_n = \frac{a(1-r^n)}{1-r}$. Plugging in our values:

$$S_8=\frac{1-(\frac12)^8}{1-\frac12}=2\left(1-\frac1{256}\right)=\frac{255}{128}.$$

Thus the sum of the series is $\frac{255}{128}$.

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Takeaways

• Spotting a geometric progression quickly lets you apply the formula rather than adding term by term.
• When the common ratio is a fraction, writing $(\frac12)^8$ and simplifying the compound fraction carefully avoids errors.
• Practising these quick sums builds fluency for more involved sequence problems in HSC exams.

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Further Readings

HSC Probability, HSC Combinatorics, HSC Sequences, HSC Differential Equations