Question

Summing a finite geometric series with a large final term

Original question: Find the sum of the following series. Round to the nearest hundredth if necessary.

9+18+36++1474569 + 18 + 36 + \ldots + 147456

Sum of a finite geometric series:

Sn=a1(1rn)1rS_n = \frac{a_1(1-r^n)}{1-r}

Answer Attempt 1 out of 3

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Expert Verified Solution

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Expert intro: This problem is a finite geometric series with a known first term, common ratio, and last term. The main goal is to determine how many terms are present, then apply the standard sum formula accurately.

Detailed walkthrough

Identify the pattern of the sequence

The series is

9+18+36++147456.9 + 18 + 36 + \ldots + 147456.

Each term is obtained by multiplying the previous term by 2, so this is a geometric sequence with

a1=9,r=2.a_1=9, \qquad r=2.

Because the final term is given, we can find the number of terms before calculating the sum.

Find the number of terms

For a geometric sequence,

an=a1rn1.a_n=a_1r^{n-1}.

Set the last term equal to 147456:

92n1=147456.9\cdot 2^{n-1}=147456.

Divide by 9:

2n1=16384.2^{n-1}=16384.

Since

16384=214,16384=2^{14},

we get

n1=14n=15.n-1=14 \quad \Rightarrow \quad n=15.

So there are 15 terms in the series.

Apply the finite geometric series formula

Use

Sn=a1(1rn)1r.S_n=\frac{a_1(1-r^n)}{1-r}.

Substitute a1=9a_1=9, r=2r=2, and n=15n=15:

S15=9(1215)12.S_{15}=\frac{9(1-2^{15})}{1-2}.

Since 12=11-2=-1, this becomes

S15=9(2151).S_{15}=9(2^{15}-1).

Now compute:

215=32768,2^{15}=32768,

so

S15=9(327681)=932767=294903.S_{15}=9(32768-1)=9\cdot 32767=294903.

Final answer

The sum of the series is

294903.\boxed{294903}.

Because the question asks for the nearest hundredth only if necessary, and this sum is an integer, no decimal rounding is needed.

💡 Pitfall guide

A frequent mistake is to try to add the terms manually without first finding the number of terms. In a geometric series, the last term is extremely useful because it lets you solve for nn quickly. Another error is using the common ratio formula with the wrong first term after spotting 1818 or 3636 first instead of starting from 9. Students also sometimes forget that the series doubles each time, so the ratio is 2, not 9 or 18.

🔄 Real-world variant

If the same series were written as 18+36+72++29491218 + 36 + 72 + \ldots + 294912, the method would be identical, but the first term would be 18 and the number of terms would still be found from the final term. A variant question might ask for the sum of only the first 10 terms of the original sequence. In that case, you would not use the last term at all; instead, you would use n=10n=10 directly in the geometric series formula.

🔍 Related terms

common ratio, finite series sum, geometric progression

FAQ

How do you find the number of terms in a geometric series when the last term is given?

Use the nth-term formula a_n = a_1 r^(n-1), set it equal to the last term, and solve for n.

What formula gives the sum of a finite geometric series?

Use S_n = a_1(1-r^n)/(1-r) or the equivalent form a_1(r^n-1)/(r-1). Choose the version that is easiest to evaluate for the common ratio.

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