Question 1146196
Find the sum of: 
a) the first odd 'n' odd positive integers 

<pre>That is the sequence 1, 3, 5, 7, 9, …

a<sub>1</sub> = first term = 1
d = common difference = 2nd-1st = 3-1 = 2
formula for sum of first n terms:  S<sub>n</sub> = [n/2][2a<sub>1</sub>+(n-1)d]  

S<sub>n</sub> = [n/2][2a<sub>1</sub>+(n-1)d]  

S<sub>n</sub> = [n/2][2∙1+(n-1)2]

S<sub>n</sub> = [n/2][2+2(n-1)]

S<sub>n</sub> = [n/2][2+2n-2]

S<sub>n</sub> = [n/2][2n] 

S<sub>n</sub> = [n][n]

S<sub>n</sub> = n²</pre>
b) the first 'n' even positive integers<pre>
That is the sequence 2, 4, 6, 8, 10, …

a<sub>1</sub> = first term = 2
d = common difference = 2nd-1st = 4-2 = 2
formula for sum of first n terms:  S<sub>n</sub> = [n/2][2a<sub>1</sub>+(n-1)d]  

S<sub>n</sub> = [n/2][2a<sub>1</sub>+(n-1)d]  

S<sub>n</sub> = [n/2][2∙2+(n-1)2]

S<sub>n</sub> = [n/2][4+2(n-1)]

S<sub>n</sub> = [n/2][4+2n-2]

S<sub>n</sub> = [n/2][2n+2]

S<sub>n</sub> = [n/2][2(n+1)] 

S<sub>n</sub> = n(n+1)

</pre>c) the first 'n' positive integers and find 'n' if the sum is 210.<pre>

That is the sequence 1, 2, 3, 4, 5, …

a<sub>1</sub> = first term = 1
d = common difference = 2nd-1st = 2-1 = 1

          S<sub>n</sub> = [n/2][2a<sub>1</sub>+(n-1)d], 

         210 =  [n/2][2(1)+(n-1)(1)]
 
         210 =  [n/2][2+n-1]

         210 =  [n/2][n+1]

Multiply both sides by 2 to clear the fraction:

         420 = n[n+1]

         420 = n² + n

           0 = n² + n - 420

n² + n - 420 = 0
(n+21)(n-20) = 0

n+21=0;   n-20=0
   n=-21;    n=20

Discard the negative value since the number of terms is always 
a positive whole number.

Answer:  20

Edwin</pre>