This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and verifying the formula $S_n = a(1 - r^n) / (1 - r)$ by comparing manual and formula sums. For 3 + 6 + 12 + 24, $a=3$, $r=2$, $n=4$, manual sum is $3+6=9$, $+12=21$, $+24=45$. Rather than just one method, equating both confirms the formula: $3(1-16)/(1-2)=3(-15)/(-1)=45$. The derivation's telescoping shows why it matches perfectly! Choice B correctly computes 45 from both ways. Choice A might come from $n=3$ sum=21 doubled or error, but count terms: 4 here. Practice by computing small series both manually and with formula, watch for sign flips when $r>1$, and use this to build trust in the method— you're on a roll!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and using $S_n = a \frac{1 - r^n}{1 - r}$ to model decreasing values like trade-in offers. The series 400 + 400(0.85) + ... + 400(0.85)^5 has a=400, r=0.85, n=6 terms. Rather than adding, plug in: $400 \frac{1 - 0.85^6}{1 - 0.85}$, capturing the decay. Cancellation in derivation makes it ideal for $|r|<1$. Choice A correctly sets exponent 6 for 6 terms and denominator 1-0.85. Choice B uses 5, maybe forgetting the first term—count from 0 to 5 exponents. List terms to confirm n, compute carefully with decimals, and apply to real scenarios like this—you're developing versatile skills, keep it up!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula $S_n = a(1 - r^n) / (1 - r)$ to express sums without expanding all terms. Here, the series is 2 + 2(1.1) + 2(1.1)^2 + ... + 2(1.1)^9, which factors to 2 times (1 + 1.1 + ... + 1.1^9), so $a=2$, $r=1.1$, $n=10$ terms. Rather than adding manually (tedious for decimals!), we use $S_n = a(1 - r^n) / (1 - r)$, directly matching the given expression with the exponent 10 and denominator 1-1.1. The formula comes from a clever algebraic trick where most terms cancel when you subtract $r S_n$ from $S_n$—it's like magic that simplifies complex sums! Choice A correctly uses $n=10$ for the 10 terms (from exponent 0 to 9) and the standard denominator 1-r. Choice C flips the denominator to 1.1-1, which changes the sign and could lead to errors if not adjusted, but remember the standard form avoids confusion; Choice D omits the division, summing like a single term times $(r^n -1)$, which overcounts. Always count terms carefully (here 10), verify $r>1$ handles negative denominators positively, and practice rewriting in equivalent forms like $(a(r^n -1))/(r-1)$ for flexibility—you're doing great, keep exploring these variations!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula $S_n = a(1 - r^n) / (1 - r)$ through telescoping cancellation for efficient summation. A geometric series is $S_n = a + ar + ar^2 + \dots + ar^{n-1}$, and multiplying by r gives $r S_n = ar + ar^2 + \dots + ar^n$; subtracting them cancels intermediate terms. Rather than just memorizing, deriving it step-by-step builds deep insight: $S_n - r S_n = a - ar^n$, so $S_n (1 - r) = a(1 - r^n)$, then divide to get the formula. This telescoping is the key—most terms vanish, leaving a simple expression! Choice B correctly shows the derived formula with exponent n and denominator 1-r. Choice A uses n-1 in the exponent, which would be off by one term, tempting if you miscount the series starting at ar^0; remember the last term is ar^{n-1}, so r^n in the formula. Master the derivation by practicing the subtraction on paper, check $r \ne 1$ (use n a if so), and apply to small series to verify—this process empowers you to handle any geometric sum confidently!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying $S_n = a(1 - r^n) / (1 - r)$ to summation notation with $|r|<1$. The sum $ \sum_{k=0}^9 4 (1/3)^k $ = $ 4 \sum_{k=0}^9 (1/3)^k $, with inner sum $ (1 - (1/3)^{10}) / (1 - 1/3) $, n=10 terms. Rather than expanding, the formula handles it; note for infinite, it converges to $ a/(1-r) $, but here finite. Derivation's telescoping works for any n! Choice A correctly uses exponent 10 and denominator 1 - 1/3. Choice B has 9, perhaps off-by-one in indexing—k=0 to 9 is 10 terms. Confirm range (10 terms), simplify denominator (1 - 1/3 = 2/3), and compare to infinite for insight—this finite vs. infinite distinction is key, you're doing wonderfully!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to apply the formula S_n = a(1 - $r^n$) / (1 - r) to calculate the sum efficiently for a finite number of terms. A geometric series is the sum of terms where each is multiplied by a common ratio r; here, starting with 3, each term doubles (r=2), and there are 6 terms up to 96. Rather than adding them one by one, which is straightforward but good practice for verification, the formula S_n = a(1 - $r^n$) / (1 - r) simplifies it, where a=3, r=2, n=6, giving S_6 = 3(1 - 64) / (1 - 2) = 3(-63) / (-1) = 189. To see why it works, note the derivation involves writing the series, multiplying by r, subtracting, and watching terms cancel, leaving the compact formula—try it yourself to build intuition! Choice A correctly applies the parameters and computes 189, matching the manual sum (3+6=9, +12=21, +24=45, +48=93, +96=189). Choice C might tempt if you mistakenly take half or forget terms, like summing only to 48 for 93 then adding wrong, but remember to count all n=6 terms accurately. The key strategy is to identify a (first term), r (ratio), n (count terms), plug into the formula, and double-check with partial sums if possible—you've got this, and practicing builds confidence for larger series!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying the formula $S_n = a(1 - r^n) / (1 - r)$ to model real-world scenarios like savings with interest. The balance after 6 deposits is $200(1 + 1.01 + 1.01^2 + \cdots + 1.01^5)$, where each term represents a deposit's growth, with $a=200$. Actually, it's 200 times sum of geometric terms with first=1, r=1.01, n=6. Rather than listing each, the formula captures the pattern: $sum = (1 - 1.01^6)/(1 - 1.01)$, then multiply by 200. Derivation shows terms cancel nicely, perfect for compounding! Choice A correctly uses exponent 6 for n=6 terms and denominator $1-1.01$. Choice C flips numerator to $1.01^6 -1$ but keeps $1-1.01$, making it negative—remember equivalent form is $(r^n -1)/(r-1)$ for positive result. Identify the series structure (here starting at 1, not 200 as $a$), count terms accurately, and simulate with small n to verify—this modeling skill is powerful for finance, you've got it!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and adapting the formula S_n = a(1 - $r^n$) / (1 - r) for applications like bouncing ball distances. The total distance for 5 bounces is sum_${k=1}^5$ $210(0.8)^k$ = 20 sum_${k=1}^5$ $(0.8)^k$, where sum = 0.8(1 - $0.8^5$)/(1 - 0.8) since it starts from k=1, not 0. Rather than calculating each bounce, the formula handles the pattern: adjust to a=20*0.8=16, then $S=16(1-0.8^5$)/(1-0.8). Cancellation in derivation makes it reliable for decaying ratios like 0.8<1. Choice B correctly sets a=16, r=0.8, n=5 for the adjusted series. Choice A uses 20 as a but includes an extra term like sum from k=0, overestimating—remember to shift for starting at k=1! Factor out the first multiplier, confirm n by listing a few terms, and note |r|<1 means terms shrink—you're mastering these practical uses, great job!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to apply the formula $S_n = a(1 - r^n)/(1 - r)$ to sums with decimal ratios. A geometric series is the sum of terms from a geometric sequence: here it's $2, 21.1, 21.1^2, \ldots, \text{ up to } 21.1^9$ (10 terms, multiplying by 1.1 each time). Rather than adding manually, use $S_n = a(1 - r^n)/(1 - r)$, with $a=2$, $r=1.1$, $n=10$. Notice the series is $2(1 + 1.1 + 1.1^2 + \ldots + 1.1^9)$, but actually starts with $2=21.1^0$, so full formula applies directly. Choice A correctly applies the formula with $a=2$, $r=1.1$, $n=10$, giving $2(1 - 1.1^{10})/(1 - 1.1)$. Choice C uses $1 + r$ in the denominator instead of $1 - r$: that's a sign error, since for $r>1$, denominator is negative, but the formula accounts for it! The three-step strategy: (1) Identify $a$, $r$, $n$ (count terms carefully), (2) Check $r \ne 1$, (3) Substitute, handling decimals precisely. This formula shines for growth rates like interest—keep practicing, you've got this!

This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to apply the formula $S_n = a(1 - r^n) / (1 - r)$ to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: here the sequence is 3, 6, 12, 24, 48, 96 (multiply by 2 each time), the series is 3 + 6 + 12 + 24 + 48 + 96 (adding those terms). Rather than adding manually (tedious for many terms!), we use the formula $S_n = a(1 - r^n) / (1 - r)$, where a is first term, r is common ratio, and n is number of terms being summed. For this series, plug in a=3, r=2, n=6: $S_6 = 3(1 - 2^6)/(1 - 2) = 3(1 - 64)/(-1) = 3(-63)/(-1) = 189$; you can verify by adding step-by-step to build confidence. Choice A correctly applies the formula with accurate parameter identification and calculation, yielding 189. Choice C uses just the last term 96, perhaps forgetting to sum all terms—remember, the series is the sum, not a single term! The three-step geometric series strategy: (1) Identify a, r, n, (2) Check r ≠ 1 to avoid division by zero, (3) Substitute and calculate carefully, watching signs for r > 1. Understanding this formula turns tedious addition into quick computation—great job practicing!