Finite Geometric Sums from Partial Sums

Why Finite Geometric Sums Have a Short Formula

A geometric sequence changes by the same multiplicative factor at every step. Its terms can grow, shrink, alternate in sign, or become zero. Adding many such terms directly is possible, but a useful pattern lets us calculate the sum more efficiently.

The key idea is to write the sum once, multiply it by the common ratio, and subtract. Most terms then cancel. This works for a finite list of terms and does not require the terms to get smaller or the common ratio to have any particular size.

Terms, Common Ratio, and Partial Sum

Let $a$ be the first term and $r$ the common ratio. A finite geometric sequence with $n$ terms is defined by $t_1=a$ and $t_{k+1}=r t_k$ for each $k$ from 1 through $n-1$. The common ratio describes how one term is obtained from the preceding term.

Definition: The $n$-term finite geometric sum is $S_n=t_1+t_2+\cdots+t_n$, where $n$ is a positive integer and $t_1=a,\ t_{k+1}=r t_k$. For $r\ne 0$, its terms can also be written $t_k=ar^{k-1}$. The recurrence definition also covers $r=0$: in that case, the first term is $a$ and every later term is zero.

For example, if $a=3$ and $r=2$, the terms begin $3,6,12,24,\ldots$. If $a=5$ and $r=-2$, they begin $5,-10,20,-40,\ldots$. A negative ratio reverses the sign at each step.

When the term before a given term is nonzero, the common ratio can be found by dividing that term by the preceding term. In particular, if $a\ne 0$, then $r=t_2/t_1$. But division by a zero term is not defined, so a consecutive-term quotient cannot always be used to find the ratio.

Deriving the Finite Geometric-Sum Formula

Write the sum in order, then multiply the entire sum by $r$. Because multiplying each term by $r$ produces the next term, the two expressions line up with all but their first and last terms in common.

Write the partial sum.
$S_n=t_1+t_2+\cdots+t_n$.

Multiply by the ratio.
The recurrence gives $rS_n=t_2+t_3+\cdots+t_{n+1}$.

Subtract the two equations.
Every term from $t_2$ through $t_n$ cancels, leaving $S_n-rS_n=t_1-t_{n+1}$.

Use the first and next terms.
Since $t_1=a$ and $t_{n+1}=r^n a$, this becomes $(1-r)S_n=a-ar^n$.

This subtraction argument also explains why the formula works: the repeated middle terms cancel exactly, no matter whether $r$ is positive, negative, or zero. If $r\ne 1$, divide both sides by $1-r$. If $r=1$, that division is not possible; the recurrence instead says that every term equals $a$, so the sum is $na$.

Finite geometric-sum formula: For a positive integer $n$, first term $a$, and common ratio $r$, $$ S_n= \begin{cases} \displaystyle a\frac{1-r^n}{1-r}, & r\ne 1,\\[6pt] na, & r=1. \end{cases} $$ The formula for $r\ne 1$ includes $r=0$ and all negative ratios. No condition such as $|r|<1$ is needed for a finite sum.

Worked Examples

Worked Example: A Positive Ratio Greater Than One

Find the sum of the first five terms of the geometric sequence with first term $3$ and common ratio $2$.

Here, $a=3$, $r=2$, and $n=5$. Since $r\ne1$, use the finite geometric-sum formula:

S_5=3\frac{1-2^5}{1-2} =3\frac{1-32}{-1} =3\frac{-31}{-1} =93.

Check by writing and adding the five terms: $3,6,12,24,48$. Their sum is $3+6+12+24+48=9+12+24+48=21+24+48=45+48=93$, which matches the formula.

Worked Example: An Alternating Sum with a Negative Ratio

Find the sum of the first six terms when $a=5$ and $r=-2$.

Substitute $a=5$, $r=-2$, and $n=6$. Since $-2\ne1$, the formula applies:

S_6=5\frac{1-(-2)^6}{1-(-2)} =5\frac{1-64}{3} =5\frac{-63}{3} =5(-21) =-105.

Check from the terms: $5,-10,20,-40,80,-160$. Pairing consecutive terms gives $5-10=-5$, $20-40=-20$, and $80-160=-80$. Thus the total is $-5-20-80=-105$. The negative answer is reasonable: in this finite list, the negative terms have greater total magnitude than the positive terms.

Worked Example: A Negative Fractional Ratio

Find the sum of the first four terms when $a=12$ and $r=-\frac13$.

Here $a=12$, $r=-\frac13$, and $n=4$. The fourth power of the negative ratio is positive:

\left(-\frac13\right)^4=\frac1{81}.

Now apply the formula:

S_4 =12\frac{1-\frac1{81}}{1-(-\frac13)} =12\frac{\frac{80}{81}}{\frac43} =12\cdot\frac{80}{81}\cdot\frac34 =\frac{80}{9}.

Check by listing the terms and adding them:

12,\quad -4,\quad \frac43,\quad -\frac49, \qquad 12-4+\frac43-\frac49 =8+\frac{12}{9}-\frac49 =8+\frac89 =\frac{80}{9}.

The direct sum agrees with the formula.

Worked Example: A Zero Common Ratio

Find the sum of the first five terms when $a=7$ and $r=0$.

The recurrence gives $t_1=7$ and $t_{k+1}=0\cdot t_k=0$. Thus the five terms are $7,0,0,0,0$. Since $r=0\ne1$, the formula applies:

S_5=7\frac{1-0^5}{1-0} =7\frac{1}{1} =7.

Adding the terms directly gives $7+0+0+0+0=7$, confirming the result.

When the Ratio Is One

The separate case $r=1$ is essential. In that case the recurrence says each term is $1$ times the term before it, so all $n$ terms equal $a$. Therefore their sum is $a+a+\cdots+a=na$. The expression $a(1-r^n)/(1-r)$ cannot be used at $r=1$, because its denominator is zero.

Worked Example: A Ratio of One

Find the sum of the first seven terms when $a=-3$ and $r=1$.

All seven terms are $-3$. Apply the $r=1$ case:

$$ S_7=7(-3)=-21. $$

Check directly: $(-3)+(-3)+(-3)+(-3)+(-3)+(-3)+(-3)=-21$. The quotient formula would give $(-3)(1-1^7)/(1-1)=0/0$, which is undefined, not an alternative way to obtain the sum.

Common Pitfalls

Dividing by a zero term to find the ratio. Consider the geometric sequence $4,0,0,\ldots$. It is wrong to claim that the ratio can be found by dividing any term after the first by the term immediately before it: calculating $0/0$ is undefined. The ratio can be found from the second term divided by the first because the first term is nonzero: $0/4=0$. The recurrence confirms the ratio: multiplying each term by $0$ gives the next term. More generally, divide consecutive terms only when the preceding term is nonzero.

Using the quotient formula when $r=1$. Substituting $r=1$ makes the denominator $1-r=0$. Do not divide by zero or treat $0/0$ as a value. Use $S_n=na$, which follows because every term is $a$.

Dropping the sign on a negative ratio. For a negative ratio, calculate $r^n$ with the exponent before simplifying the numerator. In particular, an even power of a negative number is positive, while an odd power is negative. The examples with $r=-2$ and $r=-\frac13$ show why this sign matters.

Using the Formula Carefully

To calculate a finite geometric sum, first identify the first term $a$, the ratio $r$, and the number of terms $n$. Check that the terms really follow the same recurrence, then choose the correct case: $S_n=na$ if $r=1$, and $S_n=a(1-r^n)/(1-r)$ otherwise. Finally, check the result by listing or grouping the terms when practical.

The formula concerns exactly $n$ terms. It is valid for a large positive ratio, a negative ratio, and a zero ratio; it does not require the sequence to converge. That is a feature of the finite calculation: the derivation uses cancellation in a finite expression and then divides only when $1-r\ne0$.

Key takeaway: Multiplying a finite geometric sum by its common ratio shifts its terms by one place. Subtracting the shifted sum cancels the middle terms, giving $S_n=a(1-r^n)/(1-r)$ when $r\ne1$. When $r=1$, all terms are equal and $S_n=na$.

Check Your Understanding

Use the finite geometric-sum formula where appropriate, and show how you check each result.

Find the sum of the first five terms when $a=2$ and $r=3$.
Find the sum of the first four terms when $a=6$ and $r=-1$.
Find the sum of the first six terms when $a=9$ and $r=1$.
For the sequence $8,0,0,\ldots$, what is its common ratio, and why is dividing the third term by the second not a valid way to find it?
Does the finite geometric-sum formula require $|r|<1$? Explain what in its derivation determines the condition that is actually needed.