1)The first term of a sequence is 13. Starting with the second term, each term is the sum of the cubes of the digits in the previous term. For example, the second term is \(1^3+3^3=28\) Find the 100th term.

2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

3)Find \( 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + \dots + 2^2 - 1^2. \)

SUS101 Mar 9, 2019

#1**+1 **

1.

\(a_1 = 13\\ a_2 = 28\\ a_3 = 520\\ a_4 = 133\\ a_5 = 55\\ a_6 = 250\\ a_7 = 133\\ a_8 = 55\\ a_9 = 250\\ \vdots\)

As you see, for all \(n \geq 4\), \(a_n = \begin{cases}133&&n\equiv1\pmod3\\ 55&& n\equiv 2 \pmod 3\\ 250&& n\equiv 0 \pmod 3\end{cases}\).

\(\therefore a_{100} = 133\)

MaxWong Mar 9, 2019

#2**+1 **

2.

Assume \(a_1 = x,\quad a_2 = y\).

\(a_3 = x + y\\ a_4 = x+2y\\ a_5 = 2x+3y\\ a_6 = 3x + 5y\\ a_7 = 5x + 8y\\ a_8 = 8x + 13y\\ a_9 = 13x + 21y\\ a_{10} =21x+34y\)

WLOG, assume y = 0 \(\implies x = \dfrac{6}{5}\).

Sum of 10 terms = \(\dfrac{6}{5} \left(1+0+1+1+2+3+5+8+13+21\right) = \dfrac{6}{5} \cdot(55) = 66\)

MaxWong Mar 9, 2019

#4**0 **

Max: The last one forms an Arithmetic Sequence as follows:

199, 195, 191, 187, 183......and so on for 50 terms

Sum =50/2*[2*199 + (50 -1) * - 4]

Sum =5050

Guest Mar 9, 2019

#6**+1 **

2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

a1 = a1

a2 = a2

a3 = a1 + a2

a4 = a2 + a3 = a1 + 2a2

a5 = a3 + a4 = 2a1 + 3a2

a6 = a4 + a5 = 3a1 + 5a2

a7 = a5 + a6 = 5a1 + 8a2 = 6 → a1 = (1/5) (6 - 8a2) (1)

a8 = 8a1 + 13a2

a9 = 13a1 + 21a2

10 = 21a1 + 34a2

So....the sum of the first ten terms = 55a + 88a2

Subbing (1) into this.....the sum is

55(1/5) (6 - 8a2) + 88a2 =

11(6 - 8a2) + 88a2 =

66 - 88a2 + 88a2 =

66

CPhill Mar 9, 2019