+870 Exercise 1
Let (un) be a geometric series, which common ratio r equals 5 and 1st term u0=10.
Exercise 2
Let (un) be a sequence defined by un+1=un×\(\frac{3}{2}\), and u0=2.
Exercise 3
For each of the following sequences, conjecture if they are, or not, geometrical series;
- If they seem to be, prove it.
- If they aren't obviously, find a counterexample (just calculate the first 3 terms of the sequence).
Lesson on geometrical series: http://www.regentsprep.org/regents/math/algtrig/ATP2/GeoSeq.htm
+129852 Here's the first one :
1) 10, 50, 250, 1250, 6250, 31250, 156250, 781250, 3906250, 19531250
2) u50 = 10*5(50 - 1) = 10*549 = 177635683940025046467781066894531250
3) Sum of the first 49 terms = u0 * [ 1 - rn ] / [ 1 - r ] = 10 * [ 1 - 549] / [ 1 - 5] =
44408920985006261616945266723632810
+129852 Here's the first one :
1) 10, 50, 250, 1250, 6250, 31250, 156250, 781250, 3906250, 19531250
2) u50 = 10*5(50 - 1) = 10*549 = 177635683940025046467781066894531250
3) Sum of the first 49 terms = u0 * [ 1 - rn ] / [ 1 - r ] = 10 * [ 1 - 549] / [ 1 - 5] =
44408920985006261616945266723632810
+870 Thanks CPhill, your answer is exact.
You got
for the first exercise, plus a brownie:
You'll get some more if you find the answer for the two other exercises too. 
+870 Thanks CPhill, your answer is exact.
You got
for the first exercise, plus a brownie:
You'll get some more if you find the answer for the two other exercises too.
+870 Thanks CPhill, your answer is exact.
You got
for the first exercise, plus a brownie:
You'll get some more if you find the answer for the two other exercises too.
+870 And excuse me, I don't understand why my reply was posted three times instead of just once...
+129852 Here's my attempt at the second
1) It is geometric the first few terms are 2, 3, 9/2, 27/4...... the generating "formula" for any term is : 2*(3/2)(n - 1)
2) We can prove this by dividing the (n+ 1)th term by the nth term....we have :
[2(3/2) ((n - 1) + 1) ] / [ 2 (3/2)n-1 ] = (3/2)n / (3/2)n-1 = 3/2 = r = the common difference between terms
+870 Correct again.
for the second exercise.
Ah, and I almost forgot the brownie
+129852 Here's my best guesses for the last one :
1) 4n ..... not geometric ...... the first few terms are : 4, 8, 12, 16.......
This is an arithmetic series, instead
2) n4 .....not geometric ......the first few terms are : 1, 16, 81, 256
Note that the geometric difference between the first and second terms = 16, but between the 2nd and 3rd terms is only 81 / 16 .....geometric series always have a constant difference between terms
3) 3n / 5 n + 1 let's examine the first few terms before making any speculation....we have.....
3/25, 9/125/, 27/ 625........
The ratio of the second term to the first = [9/ 125] /[ [3 / 25] = 3/5
The ratio of the 3rd to the second = 3/5
So....it appears to be geometric ....proof....
[ 3 n + 1 / 5 n + 2 ] / [3n / 5 n + 1 ] = [ 3 n + 1 / 5 n + 2 ] * [ 5 n + 1 / 3n ] = 3 / 5
4) 3 + 3n .....here's the first few terms......
6, 9, 12, 15......... this is obviously not geometric........the ratio between the 2nd term to the first = 3/2 ...but the ratio between the 3rd and 2nd = 12/9 = 4/3
+870 This is exact. Your marks :
| Exercise | Mark |
| Ex.1 | 20/20 |
| Ex.2 | 20/20 |
| Ex.3 | 20/20 |
| Average | 20/20 |
| TOTAL | 60/60 |
You deserved a chocolate mufin for such a good report card !
P.S.: If you don't like chocolate or prefer another flavor, tell me, I'll find it.
