Asked by Karen Paula on Apr 28, 2024
Verified
Find the partial sum. ∑i=142(32) i−1\sum_{i=1}^{4} 2\left(\frac{3}{2}\right) ^{i-1}∑i=142(23) i−1
A) 1054\frac{105}{4}4105
B) 174\frac{17}{4}417
C) 654\frac{65}{4}465
D) 894\frac{89}{4}489
E) 794\frac{79}{4}479
Partial Sum
The sum of a portion of a sequence of terms. It is often used in reference to the sum of the first \( n \) terms of an infinite series.
- Learn to conceptualize and perform calculations of partial sums in geometric progressions.
Verified Answer
ZK
Zybrea Knight
May 03, 2024
Final Answer :
C
Explanation :
Using the formula for the sum of a finite geometric series with first term $a$ and common ratio $r$, we have:
∑i=142(32)i−1=2(32)0+2(32)1+2(32)2+2(32)3=2+3+4.5+6.75=654\begin{align*}\sum_{i=1}^{4} 2\left(\frac{3}{2}\right)^{i-1} &= 2\left(\frac{3}{2}\right)^0 + 2\left(\frac{3}{2}\right)^1 +2\left(\frac{3}{2}\right)^2 +2\left(\frac{3}{2}\right)^3\\&= 2+3+4.5+6.75 \\&= \frac{65}{4}\end{align*}i=1∑42(23)i−1=2(23)0+2(23)1+2(23)2+2(23)3=2+3+4.5+6.75=465
Therefore, the best choice is $\boxed{\textbf{(C)}}$.
∑i=142(32)i−1=2(32)0+2(32)1+2(32)2+2(32)3=2+3+4.5+6.75=654\begin{align*}\sum_{i=1}^{4} 2\left(\frac{3}{2}\right)^{i-1} &= 2\left(\frac{3}{2}\right)^0 + 2\left(\frac{3}{2}\right)^1 +2\left(\frac{3}{2}\right)^2 +2\left(\frac{3}{2}\right)^3\\&= 2+3+4.5+6.75 \\&= \frac{65}{4}\end{align*}i=1∑42(23)i−1=2(23)0+2(23)1+2(23)2+2(23)3=2+3+4.5+6.75=465
Therefore, the best choice is $\boxed{\textbf{(C)}}$.
Learning Objectives
- Learn to conceptualize and perform calculations of partial sums in geometric progressions.
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