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...