Question 84018
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
{{{2*highlight(100)+1=201}}} So the 101st term is 201



c)
Using the sum of arithmetic series formula:
{{{s=(n/2)*(a[1]+a[n])}}} a[1]=first term, a[n]=nth term (ending term which is the 20th term), and n is the number of terms
{{{s=(20/2)*(1+39)}}} Plug in values
{{{s=(10)*(40)}}}Simplify
{{{s=400}}} So the sum of the first 20 terms is 400.



d)
Again using the same formula
{{{s=(n/2)*(a[1]+a[n])}}}  a[1]=first term, a[n]=nth term (ending term which is the 30th term), and n is the number of terms
{{{s=(30/2)*(1+59)}}} Plug in values
{{{s=(15)*(60)}}}Simplify
{{{s=900}}}



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 {{{n^2}}}