Deriving the Geometric Partial-Sum Formula

From a Finite Pattern to a Formula

A geometric sum has terms formed by repeatedly multiplying by the same ratio. For example, \(3+6+12+24\) begins with \(3\) and multiplies each term by \(2\) to obtain the next. A finite list like this can be added term by term, but that approach becomes cumbersome when there are many terms.

The useful feature is that multiplying the whole sum by its common ratio reproduces most of its terms, shifted by one position. Subtracting the shifted sum cancels those repeated terms and leaves only two boundary terms. This gives a compact formula for any finite number of terms.

Definition: Let \(a\) be the first term and \(r\) the common ratio. The geometric partial sum with \(N\) terms is \(S_N=a+ar+ar^2+\cdots+ar^{N-1}=\sum_{k=0}^{N-1}ar^k\), where \(N\in\mathbb{N}\). The first exponent is \(0\), so the first term is \(a\).

Keeping track of the number of terms matters. In this convention, the last exponent is \(N-1\), not \(N\). The formula below uses exactly that convention.

Deriving the Finite Geometric-Sum Formula

Start with \(S_N\), then multiply both sides by \(r\). Each power in the scaled sum increases by one. Subtract the scaled equation from the original: the terms from \(ar\) through \(ar^{N-1}\) cancel.

Write the partial sum.
\(S_N=a+ar+ar^2+\cdots+ar^{N-1}\).

Multiply by the ratio.
\(rS_N=ar+ar^2+\cdots+ar^{N-1}+ar^N\).

Subtract and solve.
The interior terms cancel, leaving \((1-r)S_N=a-ar^N\). If \(r\ne1\), divide by \(1-r\).

\begin{aligned} S_N&=a+ar+ar^2+\cdots+ar^{N-1},\\ rS_N&=ar+ar^2+\cdots+ar^{N-1}+ar^N,\\ S_N-rS_N&=a-ar^N,\\ (1-r)S_N&=a(1-r^N). \end{aligned}

Formula: For a geometric partial sum with \(N\) terms and common ratio \(r\ne1\), \(S_N=\dfrac{a(1-r^N)}{1-r}\). If \(r=1\), every term equals \(a\), so \(S_N=Na\).

The condition \(r\ne1\) is necessary for the displayed quotient: when \(r=1\), its denominator is zero. The sum itself is still defined, and the separate expression \(Na\) gives its value. The subtraction argument also explains why only the first and the newly introduced last term remain: all the terms between them occur in both lines.

Worked Finite Sums

Worked Example: A Positive Ratio

Find \(3+6+12+24\) using the formula, and check the result by direct addition. The first term is \(a=3\), the common ratio is \(r=2\), and there are \(N=4\) terms. Since \(r\ne1\),

S_4=\frac{3(1-2^4)}{1-2} =\frac{3(1-16)}{-1} =\frac{3(-15)}{-1} =45.

Check by adding the original terms: \(3+6=9\), \(9+12=21\), and \(21+24=45\). The formula gives the same value.

Worked Example: A Negative Ratio

Find \(5-10+20-40+80\). Here \(a=5\), \(r=-2\), and \(N=5\). The ratio is not \(1\), so

\begin{aligned} S_5 &=\frac{5\bigl(1-(-2)^5\bigr)}{1-(-2)}\\ &=\frac{5(1-(-32))}{3}\\ &=\frac{5(33)}{3} =55. \end{aligned}

Check by direct addition: \(5-10=-5\), \(-5+20=15\), \(15-40=-25\), and \(-25+80=55\). The alternating signs are included through the powers of \(-2\); they must not be replaced by powers of \(2\) without their signs.

Worked Example: The Ratio-One Case

Find the sum of seven terms when the first term is \(4\) and the common ratio is \(1\). Each term is \(4\), so the special case gives \(S_7=7\cdot4=28\). Substitution into the general quotient would divide by \(1-1=0\), so that quotient cannot be used. Check by addition: \(4+4+4+4+4+4+4=7\cdot4=28\).

Using Partial Sums to Discuss Infinite Sums

An infinite series is not added by summing infinitely many terms in a single finite calculation. Instead, consider its sequence of finite partial sums. For a geometric series beginning with \(a\), its \(N\)-th partial sum is the finite sum \(S_N\) above. By definition, the infinite series converges to a value \(S\) if the sequence of partial sums has a finite limit \(S\) as \(N\) increases without bound.

For \(r\ne1\), the finite formula rewrites that sequence as \(S_N=\dfrac{a(1-r^N)}{1-r}\). If \(a\ne0\), this identity shows precisely how the powers \(r^N\) affect the partial sums. In particular, if \(r^N\) has a finite limit \(L\), then the partial sums have the limit \(\dfrac{a(1-L)}{1-r}\). Conversely, if the partial sums have a finite limit, then the same identity implies that \(r^N\) has a finite limit: rearranging gives \(r^N=1-\dfrac{1-r}{a}S_N\). These statements are conditional; the finite formula alone does not establish what limit the powers have.

Key takeaway: A geometric infinite series converges exactly when its sequence of partial sums converges. For \(a\ne0\) and \(r\ne1\), the finite formula makes that question equivalent to whether \(r^N\) has a finite limit. If \(r^N\to L\), the series has sum \(\dfrac{a(1-L)}{1-r}\).

Worked Example: Applying a Known Power Limit

Suppose an infinite geometric series has first term \(a=6\), common ratio \(r\ne1\), and a separately established limit \(r^N\to0\). Use the partial-sum formula to find the series sum. For each \(N\),

S_N=\frac{6(1-r^N)}{1-r}.

Taking the limit under the stated assumption gives

\lim_{N\to\infty}S_N =\frac{6(1-0)}{1-r} =\frac{6}{1-r}.

This is the sum under the given power-limit condition. Check by substituting the limiting value \(r^N=0\) into the partial-sum expression: it gives \(6(1-0)/(1-r)=6/(1-r)\), as required. The condition matters: the calculation does not prove that \(r^N\to0\) for any particular \(r\); that limit must be established separately.

The ratio-one case illustrates why it must be treated separately. When \(r=1\), the partial sums are \(S_N=Na\). If \(a\ne0\), these partial sums do not approach a finite value as \(N\) increases. When \(a=0\), all terms and all partial sums equal zero. More generally, the finite formula is a tool for analyzing an infinite series, not a substitute for checking the limiting behavior required by the definition of convergence.

A Common Subtraction Error

A frequent mistake is to multiply by \(r\) but leave the powers unchanged. For example, if \(S_N=a+ar+ar^2\), then writing \(rS_N=ra+rar+rar^2\) is not yet simplified, and treating it as \(a+ar+ar^2\) is incorrect. Multiplication changes every term: \(rS_N=ar+ar^2+ar^3\).

Subtracting the correctly shifted sum gives \(S_N-rS_N=a-ar^3\), so \((1-r)S_N=a(1-r^3)\), consistent with \(N=3\). The exponent in the final term is \(3\) because multiplying the last original term \(ar^2\) by \(r\) produces \(ar^3\). Checking the first and last terms is a reliable way to catch a shifting error.

Why the Formula Matters

The geometric partial-sum formula replaces a potentially long addition with a calculation involving the first term, the common ratio, and the number of terms. More importantly, it connects finite sums with infinite series: convergence is a question about the limit of partial sums, and the formula exposes the power \(r^N\) that controls those partial sums.

Keep the order of reasoning clear. First identify the first term, ratio, and number of terms; then use the finite formula, checking whether \(r=1\). For an infinite series, examine the limiting behavior of the partial sums rather than assuming that a compact finite expression automatically has a limit.

Check Your Understanding

Use the definitions and formula above to answer these questions.

Write the \(N\)-term partial sum of a geometric sequence with first term \(a\) and ratio \(r\), including its last exponent.
In the subtraction derivation, what terms remain after \(rS_N\) is subtracted from \(S_N\)?
Why is the quotient formula not valid when \(r=1\), and what expression gives the sum in that case?
For \(a\ne0\) and \(r\ne1\), what limit of \(r^N\) would give the geometric series sum \(a/(1-r)\)?
When analyzing an infinite series, why is it necessary to consider the sequence of partial sums?