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Worked solution

Write finite geometric sums in sigma notation

Express each series in sigma notation and find its sum: (a) 3+6+12+24+48+963+6+12+24+48+96; (b) 12+14+18+116+⋯+1512\frac12+\frac14+\frac18+\frac1{16}+\cdots+\frac1{512}.

CJC · Sequences and Series · Example 25

(a) Six doubling terms

Answer

∑j=163(2j−1)=189\sum_{j=1}^{6}3(2^{j-1})=189.

Count the terms and identify the progression

3+6+12+24+48+96=∑j=163(2j−1)\begin{aligned}&3+6+12+24+48+96\\ & =\sum_{j=1}^{6}3(2^{j-1})\end{aligned}

The source uses a=3,r=2 and six terms; j is the sigma counter, distinct from the common ratio.

Use the finite GP sum

(3)(26−1)2−1=(3)(64−1)=189\begin{aligned}&\frac{(3)(2^6-1)}{2-1}\\ & =(3)(64-1)\\ & =189\end{aligned}

All printed simplification steps are retained.

How did this part go?

(b) Nine halving terms

Answer

∑j=19(12)j=511512\sum_{j=1}^{9}(\tfrac12)^j=\frac{511}{512}.

Keep the first exponent and final denominator

12+14+18+116+⋯+1512=∑j=19(12)j\begin{aligned}&\frac12+\frac14+\frac18+\frac1{16}+\cdots+\frac1{512}\\ & =\sum_{j=1}^{9}\left(\frac12\right)^j\end{aligned}

The source uses a=1/2,r=1/2 and nine terms, ending at 1/2^9.

Apply the finite-sum formula exactly

(1/2)[1−(1/2)9]1−1/2=511512\begin{aligned}&\frac{(1/2)[1-(1/2)^9]}{1-1/2}\\ & =\frac{511}{512}\end{aligned}

This retains the source substitution and fraction.

How did this part go?

Solution notes

The source writes r as the summation counter too; j is used here to distinguish that counter from the GP common ratio. The series and bounds are unchanged.