SOLUTION: Find the sum of: a) the first odd 'n' odd positive integers b) the first 'n' even positive integers c) the first 'n' positive integers and find 'n' if the sum is 210

Algebra.Com
Question 1146196: Find the sum of:
a) the first odd 'n' odd positive integers
b) the first 'n' even positive integers
c) the first 'n' positive integers and find 'n' if the sum is 210

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Find the sum of:
a) the first odd 'n' odd positive integers
That is the sequence 1, 3, 5, 7, 9, …

a1 = first term = 1
d = common difference = 2nd-1st = 3-1 = 2
formula for sum of first n terms:  Sn = [n/2][2a1+(n-1)d]  

Sn = [n/2][2a1+(n-1)d]  

Sn = [n/2][2∙1+(n-1)2]

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

Sn = [n/2][2+2n-2]

Sn = [n/2][2n] 

Sn = [n][n]

Sn = n²

b) the first 'n' even positive integers
That is the sequence 2, 4, 6, 8, 10, …

a1 = first term = 2
d = common difference = 2nd-1st = 4-2 = 2
formula for sum of first n terms:  Sn = [n/2][2a1+(n-1)d]  

Sn = [n/2][2a1+(n-1)d]  

Sn = [n/2][2∙2+(n-1)2]

Sn = [n/2][4+2(n-1)]

Sn = [n/2][4+2n-2]

Sn = [n/2][2n+2]

Sn = [n/2][2(n+1)] 

Sn = n(n+1)

c) the first 'n' positive integers and find 'n' if the sum is 210.

That is the sequence 1, 2, 3, 4, 5, …

a1 = first term = 1
d = common difference = 2nd-1st = 2-1 = 1

          Sn = [n/2][2a1+(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


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