Brain Teaser 5: Number Sequence

A sequence of numbers U₁, U₂,…, U_n,…is given by the formula

       U_n = 3(2/3)ⁿ – 1 where n is a positive integer.

a)      Find the value of U₁, U₂ and U₃

b)      Show that sigma or summation of U_n (for n = 1 to 15) = -9.014 to 4 significant figures

c)      Prove that U_{(n + 1)} = 2(2/3)ⁿ – 1

Was at Dataran Sunway (Kota D’sara) Starbucks with Angela and this Q cropped up and we solved it pretty fast! What about you?

If before I put up the answers in a week or so, you are the 1st to post the correct answers by way of comments (with brief intro of yourself), then you win a Starbucks drink from me J.

Have fun with maths!

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From Borders' Starbucks @ The Curve, 17/5/2014 (Sat, 1.10pm):

My Proposed Solutions 

a)      To find the value of U₁, U₂ and U₃ – This is straight forward!

For U1, put n = 1 into U_n = 3(2/3)ⁿ – 1      

Therefore,                   U₁ = 3(2/3)¹ – 1 = 2 - 1

                                    U₁ = 1                            (Answer)

For U_2, put n = 2 into U_n = 3(2/3)ⁿ – 1

Therefore,                   U₂ = 3(2/3)² – 1 = 4/3 - 1

                                    U₂ = 1/3                        (Answer)

For U_3, put n = 3 into U_n = 3(2/3)ⁿ – 1

Therefore,                   U₃ = 3(2/3)³ – 1 = 8/9 - 1

                                    U₃ = -1/9                      (Answer)

b)      To show that sigma or summation of U_n (from n = 1 to 15) = -9.014 to 4 significant figures

If you use conventional approach, you will discover:

No common difference: U₂ – U₁ ≠ U₃ - U₂ (-2/3 ≠ - 4/9)

No common ratio: U₂/U₁ ≠ U₃/U₂ (1/3 ≠ - 1/3)

However, when you take a good look at U_n = 3(2/3)ⁿ – 1,

You will see that:     U_n = Geometric Term – 1…(GT = ar^(n-1) = (a/r)rⁿ)

Therefore,                 U_n = 3(2/3)ⁿ – 1  ≡ (a/r)(rⁿ) – 1

Hence,                          r = 2/3

                                  a/r = 3 or, a = 3r = 3(2/3) = 2

Therefore, sigma or summation of U_n = a(1-rⁿ)/(1-r) – n

Hence, sigma or summation first 15  U_n terms (from n=1 to 15)

                                 =  a(1-rⁿ)/(1-r) – n

Where,                  a = 2 (1^st of the GT – pl see above)

                              r = 2/3 (common ratio of the geometric sequence)

                             n = 15

Hence, sigma or summation first 15  U_n terms (from n=1 to 15)

                              =  a(1-rⁿ)/(1-r) – n

                              = 2(1 – (2/3)¹⁵)/(1 – 2/3) – 15

                              = 2(0.997716341)/(1/3) – 15

                              = 6(0.997716341) – 15

                              = 5.98629805 -15

                              = - 9.01370195

                              = - 9.014 (to 4 sf) (Shown)

c)      To prove that U_{(n + 1)} = 2(2/3)ⁿ – 1

Put n = (n+1) into U_n = 3(2/3)ⁿ – 1

Therefore,             U_{(n + 1)} = 3(2/3)^{(n + 1)} – 1

                              U_{(n + 1)} = 3(2/3)ⁿ(2/3)¹ – 1…Law of Index: b^{(m + n)} = b^m x bⁿ

                              U_{(n + 1)} = 2(2/3)ⁿ – 1 (Shown)

No one wins a Starbucks drink from me...ok, got to meet up with Karina of Garden Intl School later...

Brain Teaser 4: Find the centre and the radius of the circle with the equation x2 + y2 – 4x – 6y – 12 = 0 (or, x2 + y2 = ax + by + c, generally)

If you are given the centre of a circle as (a, b) and the radius as r (directly or indirectly), you can easily use Pythagoras Theorem to derive its equation as follows:

(x – a)² + (y – b)² = r²

If you are given equation of a circle not in the above format but in already simplified form, say

x2 + y2 – 4x – 6y – 12 = 0, would you be able to find:

1) its centre (a, b); and 

2) its radius, r?

Answer for Brain Teaser 4 will be out in a week or so. Let you try first.

Have fun with maths!

Notes:

All brain teasers in this blog are questions given to students to try and they took longer than usual to come up with solutions, and in some instances, even after 15 minutes, the student(s) just could not solve them.

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My Proposed Solution

Use: Completing-the-Square (CTS) Method:

Since, the eqn of the circle given is:  x² + y² – 4x – 6y – 12 = 0

Rearrange eqn for CTS:                   (x² – 4x) + (y² – 6y) = 12

Now, do CTS:                                  (x – 4/2)² + (y – 6/2)² = 12 + (-4/2)² +  (-6/2)²

Simplify:                                          (x – 2)² + (y – 3)² = 25

(x – a)² + (y – b)² = r² format:          (x – 2)² + (y – 3)² = 5²

(Since equation of a circle with radius r and centre (a, b) on Cartesian plane can be expressed by using Pythagoras Theorem as:   (x – a)² + (y – b)² = r² 

Therefore: centre is (2, 3) and radius is 5 units.