+14 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 \( 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + \dots + 2^2 - 1^2. \)
+9673 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 \(n \geq 4\), \(a_n = \begin{cases}133&&n\equiv1\pmod3\\ 55&& n\equiv 2 \pmod 3\\ 250&& n\equiv 0 \pmod 3\end{cases}\).
\(\therefore a_{100} = 133\)
+9673 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 = \(\dfrac{6}{5} \left(1+0+1+1+2+3+5+8+13+21\right) = \dfrac{6}{5} \cdot(55) = 66\)
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
+129850 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
