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Chapter 2: Problem 56

(a)(i) Generate the sequence of partial sums of the sequence of powers of 2 : $$2^{0}, 2^{0}+2^{1}, 2^{0}+2^{1}+2^{2}, 2^{0}+2^{1}+2^{2}+2^{3}, \ldots$$ (ii) Prove that each partial sum is 1 less than the next power of 2 . (b)(i) Generate the sequence of partial sums of the Fibonacci sequence: $$F_{0}, F_{0}+F_{1}, F_{0}+F_{1}+F_{2}, F_{0}+F_{1}+F_{2}+F_{3}, \ldots$$ (ii) Prove that each partial sum is 1 less than the next but one Fibonacci number.

Short Answer

Expert verified
The partial sum of powers of 2 is one less than the next power of 2. The partial sum of Fibonacci numbers is one less than the Fibonacci number two steps ahead.

Step by step solution

01

Understanding the Problem

The exercise involves generating and analyzing the sequence of partial sums for two separate sequences: powers of 2 and the Fibonacci sequence. For each, we need to find patterns and prove mathematical statements about these partial sums.
02

Sequence of Partial Sums of Powers of 2

The sequence given is the partial sums of powers of 2. The terms are: \[ S_1 = 2^0, \; S_2 = 2^0 + 2^1, \; S_3 = 2^0 + 2^1 + 2^2, \ldots \]For a general term, the partial sum is:\[ S_n = 2^0 + 2^1 + ulletulletullet + 2^{n-1} \]
03

Proving the Formula for Powers of 2

The formula for the sum of a geometric series applies: \[ S_n = \frac{2^n - 1}{2 - 1} = 2^n - 1 \]This shows that each partial sum is 1 less than the next power of 2, since \(2^n-1\) is exactly 1 less than \(2^n\).
04

Sequence of Partial Sums of Fibonacci Numbers

The Fibonacci sequence starts as \(F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, \ldots\) with the partial sums:\[ T_1 = F_0, \; T_2 = F_0 + F_1, \; T_3 = F_0 + F_1 + F_2, \ldots\]For a general term, the partial sum is:\[ T_n = F_0 + F_1 + F_2 + \cdots + F_{n-1} \]
05

Proving the Formula for Fibonacci Sums

A known result states that the sum of the first \(n\) Fibonacci numbers is:\[ T_n = F_{n+1} - 1 \]Each partial sum \( T_n \) is therefore 1 less than the Fibonacci number two steps ahead, \(F_{n+1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
Geometric series are a sequence of terms where each term is a constant multiple of the previous one. It's a fascinating pattern where, when you sum the series, you get some interesting results. Consider the powers of 2:
  • The sequence is: \(2^0, 2^1, 2^2, \ldots \).
  • Each term is double the previous term.
When you generate the sequence of partial sums for this geometric series, you are effectively adding these powers together. The formula for the partial sum \( S_n \) of a geometric series with the first term \( a \) and a common ratio \( r \) is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] In the case of powers of 2:
  • The first term \( a \) is \( 2^0 = 1 \).
  • The common ratio \( r \) is 2.
So, the partial sum becomes \( S_n = 2^n - 1 \), which shows each sum is one less than the next power of 2.
Fibonacci Sequence
The Fibonacci sequence is a set of numbers where each number is the sum of the two preceding ones. It starts from 0 and 1, followed by numbers like 1, 2, 3, 5, etc. Each number reflects its own unique growth pattern:
  • \(F_0 = 0, \; F_1 = 1 \)
  • Subsequent terms follow: \(F_n = F_{n-1} + F_{n-2}\)
This sequence is not only a mathematical curiosity but also appears in nature, art, and various forms of expressions. When we talk about the partial sums of the Fibonacci sequence:
  • The first partial sum \( T_1 \) is \( F_0 \).
  • The second partial sum \( T_2 \) is \( F_0 + F_1 \).
In general, the partial sum \( T_n \) is the sum of the first \( n \) Fibonacci numbers. An interesting result of this summation is that \( T_n = F_{n+1} - 1 \), meaning the sum of the first \( n \) Fibonacci numbers is one less than the \((n+1)\)-th Fibonacci number.
Partial Sums
Partial sums involve taking a sequence and summing its terms one by one up to a certain point. They help in understanding the accumulative behavior of sequences:
  • For powers of 2, the partial sum of its sequence can show how quickly its values grow.
  • For Fibonacci numbers, partial sums reveal how the sequence expands over time.
The concept of partial sums is crucial because it allows us to make sense of sequences in smaller "chunks". It can turn complex sequences into understandable amounts by summing them systematically. With geometric sequences or Fibonacci numbers, the partial sums not only showcase the beauty of these sequences but also some fascinating mathematical properties. This cumulative perspective also aids in eventually understanding infinite series, where terms extend into infinity.
Mathematical Proof
Mathematical proof is the process of demonstrating that a proposition is logically true. It ensures that our mathematical world is well-structured and reliable. With sequences and series, proofs offer clarity and certainty:
  • To prove \( S_n = 2^n - 1 \) for the partial sums of powers of 2, we used the geometric series formula.
  • The formula \( T_n = F_{n+1} - 1 \) shows the sum of Fibonacci numbers, verified by known mathematical relationships.
Proofs connect intuition with formal logic. Especially in series, they help verify patterns and formulas, ensuring that results hold true in all cases. Efficient proofs often use known theorems, identities, and logical reasoning to simplify complex ideas. By providing bridge between intuition and formal mathematics, proofs make the mathematical landscape robust and dependable.

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Most popular questions from this chapter

How many prime numbers are there in each of these sequences? (Can you identify infinitely many primes in either sequence? Can you identify infinitely many non-primes?) (a) \(1,11,111,1111,11111,111111,1111111, \ldots\) (b) \(11,1001,100001,10000001, \ldots\)

The numbers in this item are all written in base 2 . (a) Carry out the addition $$\begin{array}{r}11100 \\\\+\quad 1110 \\\\\hline\end{array}$$ without changing the numbers into their base 10 equivalents - simply by applying the rules for base 2 column addition and "carrying". (b) Carry out these long multiplications without changing the numbers into their base 10 equivalents - simply by applying the rules for base 2 column multiplication. $$\begin{array}{rrr} & 10110 \\\\\text { (i) } & \times \quad 10 \\\\\hline\end{array}$$ $$\begin{array}{rrrr} & 1110 & & 110 \\\\\text { (ii) } & \times \quad 11 & \text { (iii) } & \times 111 \\\\\hline\end{array}$$ (c) Try to add these fractions (where the numerators and denominators are numerals written in base 2 ) without changing the fractions into more familiar base 10 form. $$\frac{110}{1111}+\frac{1}{10}+\frac{1001}{1110}$$

Find the lengths of the recurring blocks for: (a) \(\frac{1}{6}, \frac{5}{6}\) (b) \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) (c) \(\frac{1}{11}, \frac{2}{11}, \frac{3}{11}, \frac{4}{11}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}, \frac{8}{11}, \frac{9}{11}, \frac{10}{11}\) (d) \(\frac{1}{13}, \frac{2}{13}, \frac{3}{13}, \frac{4}{13}, \frac{5}{13}, \frac{6}{13}, \frac{7}{13}, \frac{8}{13}, \frac{9}{13}, \frac{10}{13}, \frac{11}{13}, \frac{12}{13}\)

(a)(i) Generate the first twelve terms of the Fibonacci sequence: $$F_{0}, F_{1}, \ldots, F_{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive Fibonacci numbers. Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of \(2 .\) Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences. (b)(i) Generate the first twelve terms of the sequence of powers of 2 : $$2^{0}, 2^{1}, 2^{2}, \ldots, 2^{11}$$ (ii) Use this to generate the first eleven terms of the sequence of "differences" between successive powers of 2 . Then generate the first ten terms of the sequence of "differences between successive differences". (iii) Find an expression for the \(m^{\text {th }}\) term of the \(k^{\text {th }}\) sequence of differences.

(a) Show that any decimal \(b_{n} b_{n-1} \cdots b_{0} \cdot b_{-1} b_{-2} \cdots b_{-k}\) that terminates can be written as a fraction with denominator a power of 10 . (b) Show that any fraction that is equivalent to a fraction with denominator a power of 10 has a decimal that terminates. (c) Conclude that a fraction \(\frac{p}{q},\) for which \(H C F(p, q)=1,\) has a decimal that terminates precisely when \(q\) divides some power of 10 (that is, when \(q=2^{a} \times 5^{b}\) for some non-negative integers \(\left.a, b\right)\) (d) Prove that any fraction \(\frac{p}{q},\) for which \(H C F(p, q)=1,\) and where \(q\) is not of the form \(q=2^{a} \times 5^{b}\), has a decimal which recurs, with a recurring block of length at most \(q-1\). (e) Prove that any decimal which recurs is the decimal of some fraction. \(\triangle\)
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