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Question 83751: Use the arithmetic sequence of numbers 1, 3, 5, 7, 9,….to find the following:
a) What is d, the difference between any two consecutive terms?
b) Using the formula for the nth term of an arithmetic sequence, what is 101st term?
c) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 20 terms?
d) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms?
e) What observation can you make about these sums of this sequence (HINT: It would be beneficial to find a few more sums like the sum of the first 2, then the first 3, etc.)? Express your observations as a general formula in “n.”
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! a)
The difference is the factor between each term. So going from 1 to 3, 3 to 5, 5 to 7, you see that its adding 2 each time. To verify, pick one term and subtract the previous term from it. So lets say I choose 7: I'm going to subtract 5 from it to get a difference of 2. If I pick 5, and subtract 3, I get a difference of 2.So the difference is: d=2
b)
Using what we found earlier, I know that the sequence counts up by 2 each term. So if I'm at 1 (the 1st term) and I go to 3, this means I increase by 2 each term. If I let n=0 then the term is 1, and if I let n=1 then the term is 3. This basically tells me that the arithmetic sequence is 2n+1. To verify, simply plug in the 1st term (n=0) and you'll get 1. Plug in the 2nd term (n=1) you'll get 3, if I let n=2 I get 5, etc. If I wanted to know the 101st term, let n=100 (zero is the first term) and it comes to
 So the 101st term is 201
c)
Using the sum of arithmetic series formula:
 a[1]=first term, a[n]=nth term (ending term which is the 20th term), and n is the number of terms
 Plug in values
 Simplify
 So the sum of the first 20 terms is 400.
d)
Again using the same formula
a[1]=first term, a[n]=nth term (ending term which is the 30th term), and n is the number of terms
 Plug in values
 Simplify
e)
Sum of the first 2 terms
1+3=4
Sum of the first 3 terms
1+3+5=9
Sum of the first 4 terms
1+3+5+7=16
Sum of the first 5 terms
1+3+5+7+9=25
Sum of the first 6 terms
1+3+5+7+9+11=36
Sum of the first 7 terms
1+3+5+7+9+11+13=49
Sum of the first 8 terms
1+3+5+7+9+11+13+15=64
Sum of the first 9 terms
1+3+5+7+9+11+13+15+17=81
Sum of the first 10 terms
1+3+5+7+9+11+13+15+17+19=100
Notice how the partial sums are all perfect squares. So the sums follow the sequence 
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