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Definition of a number sequence, examples
Example 1
Write down the \latex{ 1 }st, \latex{ 2 }nd, \latex{ 3 }rd,..., \latex{n^{th}} odd positive integer.
Solution
The first odd number is \latex{ 1 }, the second is \latex{ 3 }, the third is \latex{ 5 }, in general, the \latex{n^{th}} is \latex{2n-1}. This way we have defined a number sequence where we map each positive integer to a number.
DEFINITION: A number sequence is a function whose domain is the set of positive integers and its codomain is a number set.
Its notation is \latex{(a_{n} )_{n\in\N^{+} }} or simply \latex{(a_{n} ).} The \latex{n^{th}} term of the sequence is \latex{a_{n}.}
Sometimes the domain of the sequence is considered to be the set of natural numbers, in which case there is a function value at \latex{ 0 } too. It depends on the example in question which version is more useful.
Sometimes the domain of the sequence is considered to be the set of natural numbers, in which case there is a function value at \latex{ 0 } too. It depends on the example in question which version is more useful.
Example 2
Compute the sum of the first \latex{ n } odd positive integers.
Solution I
Gain some intuition: first compute the sum for \latex{n=1,2,3,4,5:}
\latex{1=1;}
\latex{1+3=4;}
\latex{1+3+5=9;}
\latex{1+3+5+7=16;}
\latex{1+3+5+7+9=25.}
\latex{1+3=4;}
\latex{1+3+5=9;}
\latex{1+3+5+7=16;}
\latex{1+3+5+7+9=25.}
It seems clear that the sums are the square numbers. We conjecture that the sum of the first \latex{n} positive integers is \latex{n^{2}}, that is,
\latex{1+3+5+...+2n-1=n^{2}.}
The conjecture is true for \latex{n=1.} Suppose that it is true for \latex{n} and using this, we show that it is also true for \latex{n+1}:
\latex{1+3+5+...+2n+1=n^{2}+2n+1=(n+1)^{2}.}
The conjecture is verified using induction.
Solution II
Visualize the numbers the following way: \latex{ n } is visualized by a block consisting of n identical squares. For example the block belonging to \latex{ 5 } can be seen in Figure 1.
Figure 1
If we place two blocks such that they are perpendicular and have a square in common, then we get an \latex{ L }-shape consisting of \latex{2n-1} squares. For example the shape representing \latex{ 9 } is in Figure 2.
If we place the \latex{ L }-shapes belonging to the different odd numbers next to each other, then after every step we get a square. More precisely, after step \latex{ n } we get a square with sides \latex{ n } long. (Figure 3)
This proves
If we place the \latex{ L }-shapes belonging to the different odd numbers next to each other, then after every step we get a square. More precisely, after step \latex{ n } we get a square with sides \latex{ n } long. (Figure 3)
This proves
\latex{1+3+5+...+2n-1=n^{2}.}
Figure 2
Example 3
Arrange the odd positive integers in the following table. (There is one number in row \latex{ 1 } and any other row contains one more element than the one directly above.)
\latex{ 1 }
\latex{ 3 }
\latex{ 5 }
\latex{ 9 }
\latex{ 7 }
\latex{ 11 }
\latex{ 17 }
\latex{ 15 }
\latex{ 13 }
\latex{ 19 }
\latex{ 25 }
\latex{ 27 }
\latex{ 29 }
\latex{ 23 }
\latex{ 21 }
\latex{ 1 }
\latex{ n }
\latex{ n }
\latex{ 2n-1 }
\latex{ 7 }
\latex{ 5 }
\latex{ 3 }
\latex{ 9 }
Figure 3
We will look to answer the following questions regarding the table:
- How many numbers are there in the first \latex{ n } rows?
- Which number is the last element of row \latex{ n }?
- What is the sum of the elements in the first \latex{ n } rows?
- What is the sum of the elements in the \latex{n^{th}} row?
Solution (a)
There are \latex{1+2+3+...+n=\frac{n\times (n+1)}{2}} numbers in the first \latex{ n } rows.
We already know the sum of the first \latex{ n } positive numbers from our previous studies.
(Figure 4)
We already know the sum of the first \latex{ n } positive numbers from our previous studies.
(Figure 4)
Solution (b)
The last element of row \latex{ n } is the \latex{\frac{n\times (n+1)}{2}} th odd positive integer, which, according to Example \latex{ 1 }, is
\latex{2\times \frac{n\times (n+1)}{2}-1=n^{2}+n-1.}
\latex{ n-1 }
\latex{ n }
Figure 4
Solution (c)
We can apply Example \latex{ 2 } since the sum of the first \latex{\frac{n\times (n+1)}{2}} odd positive integers is
\latex{\left\lgroup\frac{n\times (n+1)}{2} \right\rgroup ^{2}.}
Solution (d)
Using (c) we can easily compute this one too. Subtracting the sum of the first \latex{n-1} rows from the sum of the first \latex{ n } rows we get the sum of the numbers in row \latex{ n }:
\latex{\left(\frac{n\times (n+1)}{2} \right)^{2}-\left(\frac{(n-1)\times n}{2} \right)^{2}=\frac{n^{2}\times ((n+1)^{2}-(n-1)^{2} ) }{4}=n^{3}.}
Thus the sum of the numbers in the \latex{n^{th}} row is the \latex{n^{th}} positive cube.
It is worth noting that, by using the result from (3), the sum of the first \latex{ n } positive cubes is
\latex{1^{3}+2^{3}+3^{3}+...+n^{3}=\left(\frac{n\times (n+1)}{2} \right)^{2}=(1+2+3+...+n )^{2}.}
(We have already proved this in the previous chapter.)
The result is illustrated in Figure 5 for \latex{n=4} .
We have already proved that by adding the first \latex{ n } terms of the sequence of positive integers we get
The result is illustrated in Figure 5 for \latex{n=4} .
We have already proved that by adding the first \latex{ n } terms of the sequence of positive integers we get
\latex{1+2+3+...+n=\frac{n\times (n+1)}{2}.}
The sum of the first \latex{ n } terms of the sequence of positive cubes is
\latex{1^{3}+2^{3}+3^{3}+...+n^{3}=\left(\frac{n\times (n+1)}{2} \right)^{2}.}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
Figure 5
Example 4
Compute the sum of the first \latex{ n } positive square numbers:
\latex{1^{2}+2^{2}+3^{2}+...+n^{2}=s_{n}\times s_{n}=?}
Solution
We use a new idea for this problem. Sum the following \latex{ n } – obviously true – equalities:
\latex{(n+1)^{3}-n^{3}=3n^{2}+3n+1;}
\latex{n^{3}-(n-1)^{3}=3\times (n-1)^{2}+3\times (n-1)+1;}
.
.
.
.
.
\latex{3^{2}-2^{2}=3\times 2^{2}+3\times 2+1;}
\latex{\frac{2^{2}-1^{2}=3\times 2^{2}+3\times 2+1 }{(n+1)^{3}-1=3\times s_{n}+3\times (1+2+...+n)+n }}
After rearranging we get
\latex{3\times s_{n}=(n+1)^{3}-(n+1)-3\times \frac{n\times (n+1)}{2}=}
\latex{=(n+1)\times \left(n^{2}+2n+1-1-\frac{3}{2}\times n \right)=n\times (n+1)\times \left(n\times \frac{1}{2} \right),}
\latex{=(n+1)\times \left(n^{2}+2n+1-1-\frac{3}{2}\times n \right)=n\times (n+1)\times \left(n\times \frac{1}{2} \right),}
from which \latex{s_{n}=\frac{n\times (n+1)\times (2n+1)}{6}.}
Example 5
On \latex{ 1 } January, we deposit \latex{ 1 } euro at Ideal Bank with \latex{ 100 }% annual interest. How much will our deposit worth at the end of the year if they compound the interest
- annually;
- monthly;
- daily
(that is, they add the interest to the deposit and from that moment on the increased amount earns interest)?
Solution (a)
Obviously, interest is \latex{ 1 } euro until the end of the year therefore the deposit will be \latex{ 1 + 1 = 2 } euros.
Solution (b)
Monthly compounding means that one month's interest is added to the deposit at the end of each month and in the next month this increased deposit will earn interest. Let \latex{t_{1},t_{2},....,t_{12}} denote the worth of the deposit at the end of each month. Then
\latex{t_{1}=1+\frac{1}{12},}
\latex{t_{2}=1+\frac{1}{12}+\left(1+\frac{1}{12} \right)\times \frac{1}{12}=\left(1+\frac{1}{12} \right)^{2},}
\latex{t_{3}=\left(1+\frac{1}{12} \right)^{2}+\left(1+\frac{1}{12} \right)^{2}\times \frac{1}{12}=\left(1+\frac{1}{12} \right)^{3},}
\latex{t_{12}=\left(1+\frac{1}{12} \right)^{12}\approx 2.61} euros.
\latex{t_{2}=1+\frac{1}{12}+\left(1+\frac{1}{12} \right)\times \frac{1}{12}=\left(1+\frac{1}{12} \right)^{2},}
\latex{t_{3}=\left(1+\frac{1}{12} \right)^{2}+\left(1+\frac{1}{12} \right)^{2}\times \frac{1}{12}=\left(1+\frac{1}{12} \right)^{3},}
\latex{t_{12}=\left(1+\frac{1}{12} \right)^{12}\approx 2.61} euros.
Solution (c)
Assume that there are \latex{ 365 } days in the year. Then when we compound each day, the deposit is increased by \latex{\frac{1}{365}} th of it. According to this, by the end of the year the deposit will be worth
\latex{\left(1+\frac{1}{365} \right)^{365}=2.7145} euros.
It follows from the above if we divide the year into n parts and the deposit is compounded after each part, then by the end of the year the deposit will be worth
\latex{\left(1+\frac{1}{n} \right)^{n} } euros.
How large can this value be for large \latex{ n }? It can be proven that
\latex{\left(1+\frac{1}{n} \right)^{n}\lt e\approx 2.71828....,}
where e is the so-called Euler's number.
e is irrational and is not a root of any polynomial with integer coefficients.
According to this the maximal profit possible for a \latex{ 1 } euro deposit with the given conditions, with continuous compounding, is \latex{ 1.72 } euros.
e is irrational and is not a root of any polynomial with integer coefficients.
According to this the maximal profit possible for a \latex{ 1 } euro deposit with the given conditions, with continuous compounding, is \latex{ 1.72 } euros.
Exercises
{{exercise_number}}. In the sequence of positive even numbers, find the \latex{n^{th}} term and compute the sum of the first \latex{ n } terms.
{{exercise_number}}. In the following triangle made out of numbers, we placed the positive integers such that \latex{ 1 } goes in the first row and there are two more numbers in each row than directly above.
Answer the following questions.
- Which number is the last element of row \latex{ n }?
- What is the sum of the elements in the first \latex{ n } rows?
- What is the sum of the elements in the \latex{n^{th}} row ?
\latex{ 1 }
\latex{ 3 }
\latex{ 2 }
\latex{ 4 }
\latex{ 8 }
\latex{ 7 }
\latex{ 6 }
\latex{ 5 }
\latex{ 9 }
\latex{ 15 }
\latex{ 16 }
\latex{ 14 }
\latex{ 13 }
\latex{ 12 }
\latex{ 11 }
\latex{ 10 }
\latex{ 18 }
\latex{ 17 }
\latex{ 19 }
\latex{ 20 }
\latex{ 21 }
\latex{ 22 }
\latex{ 23 }
\latex{ 24 }
\latex{ 25 }
{{exercise_number}}. Prove the following formulae:
- \latex{1\times 2+2\times 3+3\times 4+4\times 5+...+n\times (n+1)=\frac{n\times (n+1)\times (n+2)}{3};}
- \latex{1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...+n\times (n+1)\times (n+2)=\frac{n\times (n+1)\times (n+2)\times (n+3)}{4};}
- \latex{\frac{1}{1\times 2}+\frac{1}{1\times 2}+\frac{1}{3\times 4}+...+\frac{1}{n\times (n+1)}=1-\frac{1}{n+1};}
- \latex{\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+...+\frac{1}{n\times (n+1)\times (n+2)}=\frac{1}{2}\times \left(\frac{1}{2}-\frac{1}{(n+1)\times (n+2)} \right).}
{{exercise_number}}. Two sequences, \latex{(a_{n} )} and \latex{(b_{n} )} are given. Prove that they consist of the same terms.
- \latex{a_{n}=(n+1)\times (n+2)\times (n+3)\times ...\times (2n-1)\times 2n,} \latex{b_{n}=1\times 3\times 5\times ...\times (2n-1)\times 2^{n};}
- \latex{a_{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2n-1}-\frac{1}{2n},} \latex{b_{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}.}
{{exercise_number}}. Investigate the following equations. State a general conjecture, then prove it.
\latex{1+2=3}
\latex{4+5+6=7+8}
\latex{9+10+11+12=13+14+15}
\latex{16+17+18+19+20=21+22+23+24}
\latex{4+5+6=7+8}
\latex{9+10+11+12=13+14+15}
\latex{16+17+18+19+20=21+22+23+24}