Sum a finite geometric progression

Model the doubling reward on a chessboard

The source introduces geometric series through a legend about the inventor of chess and an emperor of India. Offered a reward, the inventor asks for 1 grain on the first square, 2 on the next, 4 on the next, 8 on the next, then doubling across all 64 squares.

The emperor agrees because the request sounds small; in the story, the treasurer returns after a week reporting a quantity beyond centuries of rice production.

These are the source’s story details, not measured production data.

The mathematical model has a=1,r=2 and 64 terms:

$$S_{64}=1+2+4+8+\cdots+2^{63}=\sum_{j=0}^{63}2^j.$$

Applying the finite formula derived next gives $2^{64}-1=18\,446\,744\,073\,709\,551\,615$ grains. This exact integer is an authored answer to the source’s question; the source itself introduces the general derivation without printing the integer. The 64th square contains $2^{63}$ grains; the total uses $2^{64}-1$, so these are different quantities.

Derive the finite geometric-sum formula

Adding the terms $a,ar,ar^2,ar^3,ar^4,ar^5,\ldots,ar^{n-1}$ gives a geometric series, whose first-n-term sum is $S_n$. Multiply that sum by r so the shared middle terms line up:

$$\begin{aligned} S_n&=a+ar+ar^2+ar^3+ar^4+ar^5+\cdots\\ &\quad+ar^{n-2}+ar^{n-1}\quad\text{(1)},\\ rS_n&=ar+ar^2+ar^3+ar^4+ar^5+ar^6+\cdots\\ &\quad+ar^{n-1}+ar^n\quad\text{(2)}. \end{aligned}$$

The explicit lists illustrate a sufficiently long sum; include each actual term once when n is small. Subtracting (2) from (1) cancels the shared powers:

$$\begin{aligned} S_n-rS_n&=a-ar^n,\\ S_n(1-r)&=a(1-r^n),\\ S_n&=\frac{a(1-r^n)}{1-r}\qquad(r\ne1). \end{aligned}$$

Subtracting in the opposite order gives the equivalent form

$$S_n=\frac{a(r^n-1)}{r-1}\qquad(r\ne1).$$

The source asks why we usually choose the second form for r>1 and the first for r<1. The respective denominator is positive, making signs easier to track.

Both forms are algebraically valid for any allowed r≠1, including negative ratios; their numerators and denominators change sign together.

At r=1 the series is constant and $S_n=na$, so do not divide by zero.

The source’s GP convention still excludes zero terms and zero ratio.

Self-Practice 7 uses a=0.6,r=1.5,n=15. Its answer 524 is rounded:

$$S_{15}=\frac{0.6(1.5^{15}-1)}{1.5-1} =\frac{42948417}{81920}\approx524.272668457\approx524$$

to three significant figures. A finite sum is valid even though r>1; no convergence assumption is needed.

Apply and compare finite geometric sums

Example 24 takes finite partial sums of the infinite list $1,2,4,8,16,\ldots$. The source derives

$$S_n=2^0+2^1+2^2+\cdots+2^{n-1} =\frac{(2^0)(2^n-1)}{2-1}=2^n-1.$$

Its strict threshold is

$$S_n>2019\quad\Longrightarrow\quad2^n-1>2019 \quad\Longrightarrow\quad2^n>2020 \quad\Longrightarrow\quad n>\frac{\ln2020}{\ln2}\approx10.980.$$

The least integer is 11; indeed $S_{10}=1023$ and $S_{11}=2047$. This is a question about a finite partial sum, not a claim that the entire infinite series has a finite total.

Example 25 links each finite GP to sigma notation. Its six doubling terms have a=3,r=2:

$$3+6+12+24+48+96=\sum_{j=1}^{6}3(2^{j-1}) =\frac{(3)(2^6-1)}{2-1}=(3)(64-1)=189.$$

Its nine halving terms have a=1/2,r=1/2:

$$\frac12+\frac14+\frac18+\frac1{16}+\cdots+\frac1{512} =\sum_{j=1}^{9}\left(\frac12\right)^j =\frac{(1/2)[1-(1/2)^9]}{1-1/2}=\frac{511}{512}.$$

Here j renames the source’s summation counter r to distinguish it from the common ratio. The source’s final comparison recommends the appropriate AP or GP formula when sigma notation describes one of those special series:

First identify the term pattern and count; sigma notation alone does not decide which formula applies.