Consider the geometric series with first term -4 and common ratio -1/2. (a) Write the general term. (b) Write the series in sigma notation. (c) Calculate the sum to infinity.
|Mathematics
Consider the geometric series with first term -4 and common ratio -1/2. (a) Write the general term. (b) Write the series in sigma notation. (c) Calculate the sum to infinity.
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Answer
Tn=−4(−21)n−1
3.1.1 Write down the general term of this series.
Step 1: Identify the first term (a) and the common ratio (r).
The first term is a=−4.
The common ratio is r=−42=−21.
Step 2: Write the general term using the formula Tn=arn−1.
Tn=−4(−21)n−1
The general term is \boxed{T_n = -4\left(-\frac{1{2}\right)^{n-1}}}.
3.1.2 Write the series in sigma notation.
Step 1: Determine the number of terms (n) in the series.
The last term is 321. Set Tn=321:
−4(−21)n−1=321(−21)n−1=32×(−4)1(−21)n−1=−1281
Since 128=27, we can write:
(−21)n−1=(−21)7n−1=7n=8
There are 8 terms in the series.
Step 2: Write the series in sigma notation.
∑k=18−4(−21)k−1
The series in sigma notation is k=18−4(−21)k−1∑.
3.1.3 Calculate the sum to infinity of this series.
Step 1: Check if the sum to infinity exists.
The common ratio is r=−21. Since ∣r∣=−21=21<1, the sum to infinity exists.
Step 2: Use the formula for the sum to infinity, S∞=1−ra.
S∞=1−(−21)−4S∞=1+21−4S∞=23−4S∞=−4×32S∞=−38
The sum to infinity is \boxed{-\frac{8{3}}}.
3.2.1 Determine S5.
Given the formula for the sum of n terms: Sn=32−32(21)n.
Substitute n=5 into the formula:
S5=32−32(21)5S5=32−32(321)S5=32−1S5=31
The value of S5 is 31.
3.2.2 How many terms must be added for the sum to be equal to 8255?
Set Sn=8255:
32−32(21)n=825532−8255=32(21)n8256−255=32(21)n81=32(21)n8×321=(21)n2561=(21)n
Since 256=28, we have:
(21)8=(21)nn=8
The number of terms that must be added is 8.
3.2.3 If 2n=p, determine the value of S5−n−S5+n in terms of p.
Step 1: Rewrite Sn in terms of 2n.
Sn=32−32(21)n=32−2n32
Given 2n=p, so Sn=32−p32.
Step 2: Determine S5−n in terms of p.
S5−n=32−32(21)5−nS5−n=32−32⋅25−n1S5−n=32−32⋅25⋅2−n1S5−n=32−32⋅322nS5−n=32−2n
Substitute 2n=p:
S5−n=32−p
Step 3: Determine S5+n in terms of p.
S5+n=32−32(21)5+nS5+n=32−32⋅25+n1S5+n=32−32⋅25⋅2n1S5+n=32−32⋅32⋅2n1S5+n=32−2n1
Substitute 2n=p:
S5+n=32−p1
Step 4: Calculate S5−n−S5+n.
S5−n−S5+n=(32−p)−(32−p1)S5−n−S5+n=32−p−32+p1S5−n−S5+n=−p+p1S5−n−S5+n=p1−p2
The value of S5−n−S5+n in terms of p is \boxed{\frac{1 - p^2{p}}}.
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3.1.1 Write down the general term of this series. Step 1: Identify the first term (a) and the common ratio (r).
Consider the geometric series with first term -4 and common ratio -1/2. (a) Write the general term. (b) Write the series in sigma notation. (c) Calculate the sum to infinity.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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3.1.1 Write down the general term of this series. Step 1: Identify the first term (a) and the common ratio (r). The first term is a = -4. The common ratio is r = (2)/(-4) = -(1)/(2). Step 2: Write the general term using the formula T_n = ar^n-1. T_n = -4(-(1)/(2))^n-1 The general term is T_n = -4(-(1)/(2))^n-1. 3.1.2 Write the series in sigma notation. Step 1: Determine the number of terms (n) in the series. The last term is (1)/(32). Set T_n = (1)/(32): -4(-(1)/(2))^n-1 = (1)/(32) (-(1)/(2))^n-1 = (1)/(32 × (-4)) (-(1)/(2))^n-1 = -(1)/(128) Since 128 = 2^7, we can write: (-(1)/(2))^n-1 = (-(1)/(2))^7 n-1 = 7 n = 8 There are 8 terms in the series. Step 2: Write the series in sigma notation. _k=1^8 -4(-(1)/(2))^k-1 The series in sigma notation is _k=1^8 -4(-(1)/(2))^k-1. 3.1.3 Calculate the sum to infinity of this series. Step 1: Check if the sum to infinity exists. The common ratio is r = -(1)/(2). Since |r| = |-(1)/(2)| = (1)/(2) < 1, the sum to infinity exists. Step 2: Use the formula for the sum to infinity, S_ = (a)/(1-r). S_ = (-4)/(1 - (-1)2) S_ = (-4)/(1 + 1)2 S_ = (-4)/(3)2 S_ = -4 × (2)/(3) S_ = -(8)/(3) The sum to infinity is -(8)/(3). 3.2.1 Determine S_5. Given the formula for the sum of n terms: S_n = 32 - 32((1)/(2))^n. Substitute n=5 into the formula: S_5 = 32 - 32((1)/(2))^5 S_5 = 32 - 32((1)/(32)) S_5 = 32 - 1 S_5 = 31 The value of S_5 is 31. 3.2.2 How many terms must be added for the sum to be equal to (255)/(8)? Set S_n = (255)/(8): 32 - 32((1)/(2))^n = (255)/(8) 32 - (255)/(8) = 32((1)/(2))^n (256 - 255)/(8) = 32((1)/(2))^n (1)/(8) = 32((1)/(2))^n (1)/(8 × 32) = ((1)/(2))^n (1)/(256) = ((1)/(2))^n Since 256 = 2^8, we have: ((1)/(2))^8 = ((1)/(2))^n n = 8 The number of terms that must be added is 8. 3.2.3 If 2^n = p, determine the value of S_5-n - S_5+n in terms of p. Step 1: Rewrite S_n in terms of 2^n. S_n = 32 - 32((1)/(2))^n = 32 - (32)/(2^n) Given 2^n = p, so S_n = 32 - (32)/(p). Step 2: Determine S_5-n in terms of p. S_5-n = 32 - 32((1)/(2))^5-n S_5-n = 32 - 32 · (1)/(2^5-n) S_5-n = 32 - 32 · (1)/(2^5 · 2^-n) S_5-n = 32 - 32 · (2^n)/(32) S_5-n = 32 - 2^n Substitute 2^n = p: S_5-n = 32 - p Step 3: Determine S_5+n in terms of p. S_5+n = 32 - 32((1)/(2))^5+n S_5+n = 32 - 32 · (1)/(2^5+n) S_5+n = 32 - 32 · (1)/(2^5 · 2^n) S_5+n = 32 - 32 · (1)/(32 · 2^n) S_5+n = 32 - (1)/(2^n) Substitute 2^n = p: S_5+n = 32 - (1)/(p) Step 4: Calculate S_5-n - S_5+n. S_5-n - S_5+n = (32 - p) - (32 - (1)/(p)) S_5-n - S_5+n = 32 - p - 32 + (1)/(p) S_5-n - S_5+n = -p + (1)/(p) S_5-n - S_5+n = (1 - p^2)/(p) The value of S_5-n - S_5+n in terms of p is (1 - p^2)/(p). 3 done, 2 left today. You're making progress.