Problems about Sequences

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Problems about Sequences

READ THIS FIRST

These exercises are based on the theory treated on the page Sequences and limits

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Sequences

Level 1 problems

The angles of a rectangular triangle are the terms of an arithmetic sequence. Calculate these angles.

The angles are (pi/2 - 2h, pi/2 - h, pi/2).
The sum of these angles is 3.(pi/2) - 3.h = pi .
Hence h = pi/6.
The angles are (pi/6, 2pi/6, pi/2) .

In an arithmetic sequence is t(2) = 3.t(3) .
The sum of n terms, starting from t(1), is 0. Calculate n.

 
The sequence is a, a+v, a+2v, ...

a + v = 3.(a + 2v)  <=> 2a + 5v = 0 <=> 2a = -5v

The n-th term is t(n)=a + (n-1)v

The sum of the first n terms is (a + a + (n-1)v).n/2 = 0

<=> (2a + (n-1)v)=0

So,  -5v + (n-1)v = 0  and since v is not 0, we have n = 6.

 
 Calculate
        1 + x + x² + x³ +  ...     + x^n-1
        -----------------------------------------
        1 + x² + x⁴ + x⁶ +  ...     + x^2n-2

 
The numerator is the sum of n terms of a geometric sequence
with ratio x.
This sum is
        1.(xⁿ -1)
        -----------                     (1)
         (x - 1)

The denominator is the sum of n terms of a geometric sequence
with ratio x² .

This sum is
        1.(x²ⁿ - 1)
        --------------                  (2)
         (x² - 1)

From (1) and (2), we have that the given fraction =
         (xⁿ -1)     (x²ⁿ - 1)       (xⁿ -1).(x+1)
        --------- : ------------   = --------------
         (x - 1)      (x²  - 1)         (x²ⁿ - 1)

The sides of a triangle form a geometric sequence. What are the limits of the ratio.
Let a, aq, aqq be the sides of the triangle in ascending order. The condition for the sides is
```
 
        aq²< a + aq
<=>
         q²< 1 +  q
<=>
         q² - q - 1 < 0
<=>
        q in [1, (1 + sqrt(5))/2 [
```

The points with coordinates (a,b) (a',b') (a",b") are points of a parabola y = 3x² .
The numbers a, a', a" constitute an arithmetic sequence and b,b',b" form a geometric sequence.
Calculate the ratio of the geometric sequence.

 
Let a = x - h  a' = x  a" = x + h

Then b = 3.(x-h)²   b' = 3.x²  b" = 3.(x+h)²

The condition for geometric sequence gives

        (3.x²)² = 9.(x-h)².(x+h)²
                ...
                h = sqrt(2).x
Then the ratio is

        x²
    ------------- = ... = 3 + 2.sqrt(2)
     (x-h)(x-h)

Prove that:
if a,b,c form an arithmetic sequence, then
b² + bc + c², c² + ca + a², a² + ab + b² form an arithmetic sequence.

 
   a,b,c form an arithmetic sequence  <=> 2b  =  a + c
Well,

   b² + bc + c², c² + ca + a², a² + ab + b² form an arithmetic sequence
<=>
   2(c² + ca + a²) = b² + bc + c² +  a² + ab + b²
<=>
   c² + 2ac + a² = 2b² + b(a + c)
<=>
   2c² + 4ac + 2a² = 4b² + 2b(a + c)  and since 2b  =  a + c
<=>
   2c² + 4ac + 2a² = (a + c)² + (a + c)²
<=>
   2c² + 4ac + 2a² = 2 (a + c)²
<=>
   2c² + 4ac + 2a² = 2c² + 4ac + 2a²

Level 2 problems

An arithmetic sequence has terms t(1),t(2),t(3),...
The first term t(1) = a and the common difference is v (not 0).
The terms t(5),t(9) and t(16) form a three term geometric sequence with common ratio q.
Calculate q. Calculate t(k) in terms of k and a.

 
t(k) = a + (k-1).v              (1)
t(5) = a + 4v
t(9) = a + 8v
t(16) = a + 15v

        t(5),t(9) and t(16) form a three term geometric sequence

       a + 4v        a + 8v
<=>  ----------- = ---------- = q
       a + 8v       a + 15v

<=>  a² + 19av + 60v² = a² + 16av + 64v²

<=>  3av = 4vv
                Since v is not 0

<=>  3a = 4v  <=> v = 3a/4

Combining this with (1) yields

t(k) = a + (k-1).3a/4

     = (1 + 3k)a/4

Hence t(5) = 4a and t(9) = 7a . So, q = 7/4

 
Prove that for each integer n  > 0
        1.5 + 2.5²+ 3.5²+ ... + n.5ⁿ= (5 + (4n-1)5ⁿ⁺¹)/16

 
We'll prove this by complete induction.
a) The property is trivial for n = 1
b) Assume the property is true for n = k.

        1.5 + 2.5²+ 3.5²+ ... + k.5^k= (5 + (4k-1)5^k+1)/16

Then we have to prove that the property is true for n = k+1.
        1.5 + 2.5²+ 3.5²+ ... + k.5^k+ (k+1).5^k+1

      = (5 + (4k-1)5^k+1)/16 + (k+1).5^k+1

      = (5 + (4k-1)5^k+1)/16 + 16.(k+1).5^k+1 /16

      = (5 + (4k-1)5^k+1 +  16.(k+1).5^k+1) /16

      = (5 + (20k+15)5^k+1) /16

      = (5 + (4(k+1)-1)5^k+2) /16

u₁, u₂, u₃, ... is an arithmetic sequence and u₂ - u₁ = v
S₁ = u₁ + u₂ + ... + u₉ + u₁₀
S₂ = u₁₁ + u₁₂ + ... + u₁₉ + u₂₀
S₃ = u₂₁ + u₂₂ + ... + u₂₉ + u₃₀
etc ...
Show that the sequence S₁, S₂, S₃, ... is an arithmetic sequence.

We take two consecutive terms T1 and T2 of the sequence S₁, S₂, S₃, ...

 
   T1 = u_k+1  +  u_k+2 + ...  + u_k+10
   T2 = u_k+11 + u_k+12 + ...  + u_k+20
--------------------------------------
T2-T1 = 10v  +   10v  + ...  + 10v   = 100v

Since the difference between two consecutive terms T1 en T2 of the sequence S₁, S₂, S₃, ...
is constant, the sequence is an arithmetic sequence.

Level 3 problems

Calculate the sum of the squares of the first n strictly positive integers.

 
Let S(p) = 1^p + 2^p + 3^p + ... + n^p .

(x + 1)³ = x³ + 3x²+ 3x + 1

Now, make x resp. 0 ; 1 ; 2 ; 3 ; 4 ; ...

1³ =                + 1

2³ = 1³ + 3.1²+ 3.1 + 1

3³ = 2³ + 3.2²+ 3.2 + 1

4³ = 3³ + 3.3²+ 3.3 + 1

...

(n + 1)³ = n³ + 3n²+ 3n + 1

Adding all these sums together, we have

(n + 1)³ = 3.S(2) + 3.S(1) + (n+1)

Since S(1) = n.(n+1)/2

(n + 1)³ = 3.S(2) + 3.n.(n+1)/2  + (n+1)

3.S(2) =(n + 1)³ - 3.n.(n+1)/2  - (n+1)


3.S(2) =((n + 1)² - 3.n/2  - 1 ).(n+1)


3.S(2) =(2.(n + 1)²- 3.n - 2 ).(n+1)/2

3.S(2) =(2n²+ n).(n+1)/2

3.S(2) =n.(n+1)(2n+1)/2

S(2) =n.(n+1)(2n+1)/6

Calculate all m values such that the roots of the following equation constitute an arithmetic sequence.
x⁴ - (3m + 1) x² + m² = 0 (1)
Say r is a strictly positive root of the equation (1), then (-r) is a root too. Since the four roots form an arithmetic sequence, we can write these roots as -3r, -r, r, 3r.
The only monic equation with these roots is
```
 
        (x + 3r) (x + r) (x - r) (x - 3r) = 0
 <=>
        x⁴ - 10 r² x² + 9 r⁴ = 0
```
This equation is identical to (1) if and only if
```
 
(3m + 1) = 10 r²  and  9 r⁴ = m²
```
1. m = 3 r²
  Then 9r² + 1 = 10 r² <=> ... <=> m = 3
2. m = - 3 r²
  Then - 9r² + 1 = 10 r² <=> ... <=> m = -3/19

In a sequence is the sum of the first n terms = s_n = 2^n.p - 1, with p = a fixed real number.

Show that the sequence is geometric.

 
t_n = s_n+1 - s_n = 2^{n.p + p} - 1 - 2^n.p + 1

    = 2^{n.p + p} - 2^n.p

    = 2^n.p.(2^p - 1)

t_n-1 = 2^{n.p - p}.(2^p - 1)

t_n / t_n-1 = 2^p = independent of n = constant.

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