Calculus 2 Geometric

Question

Consider the sequence (an)n≥0 that starts 9,21,33,45,... 

What is the next term in the sequence?

Find the sum of the first 100 terms of the sequence:

Find a formula for the nth term of the sequence.

 

Answer 1

9,21,33, 45, 57, 69, ...

it's an arithmetic sequence with common difference = 12

the nth term, an,

= 9+(n-1)(12)

= 9+12n-12

= 12n-3

the 5th term = a5 = 12(5)-3 =60-3 = 57

the 100th term = a100 = 12(100)-3 = 1197

sum of 1st n integers = n(n+1)/2 by Gaus's formula

manipulate that formula to get the 1st 100 terms of 9+21+...+12n-3+ ... 1185 +1197

sum of 1 through 1197 = 100(101)/2 = 50(101) = 5050

manipulate that number to get 9+21+..+1197

12n-3

12(5050)-3(100) = 60,300

1st & last = 9 +1197= 1206

2nd&2nd to last = 21+1185 = 1206

there are 100/2 = 50 such pairs each =1206

50x1206

= 60,300= sum of the arithmetic sequence from a1 to a100, from 9 to 1197