find the sums of the following series:

Question

a) 5+7+9+....+75

 

 

 

b) (x+1)+(2x+1)+(3x+1)+....+(21x+1)

Answer 1

The key to deriving a formula to figuring out the series is to add the first and last term, the second and second to last term, and so on. Usually, you'll see a pattern in doing this.

 

For part a) The series becomes (5+75) + (7+73) + (9+71) + … + 80 = n/2*80 Note: n is the total number terms added in the series, and when you add the first and last term, add the second and second to last term, etc. then you get half the terms. To find out what n is, you just need to get an equation for the sequence in the series in which is:

 

2n+3=75 in which implies n =36 terms. Now, substitute n in the original to find the original sum. Hence,

 

36/2*80=18*80=1440

 

b) do the same as in part a)

Answer 2

a) 5+7+9+...+75

use the formula S_n=(n/2)[2a+(n-1)d] where n=number of terms, a=1st term, d=common difference between terms

the numbers 1-76 give you 38 even and 38 odd numbers

you don't have 1 or 3 so you have 36 odd numbers

S₃₆=(36/2)[2*5+(36-1)2]

   =(18)[10+(35)2]

   =(18)[10+70]

   =(18)(80)

   =1440

here's another way to look at this problem using logic and arithmetic...

there are 36 odd numbers

add the first and last: 5+75=80

add the second and next to last: 7+73=80

add the third and the next to next to last: 9+71=80

each time you add you get 80

how many groups of 80 are there ?

36 numbers grouped by twos give you 18 groups of 80

18*80=1440 again

 

b) (x+1)+(2x+1)+3x+1)+...+(21x+1)

there are 21 terms

add the x's and add the 1's

to add the x's all we have to do is add 1+2+3+...+21 and put an x after the sum

S₂₁=(21/2)[2*1+(21-1)1]

    =(21/2)[2+20]

    =(21/2)[22]

    =(21/2)(22)

    =21*11

    =231

also you could use the following: the sum of positive integers from 1 to n is n(n+1)/2

n(n+1)/2=21(22)/2=21*11=231 again

there are 21 1's which sum to 21

the answer is 231x+21, or if you want to factor out a 21 you get 21(11x+1)

Answer 3

a)

 

a₁ = 5

d = 2

n is the number of terms

 

 

 

a_n = a₁ + (n-1)d

 

75 = 5 + (n-1)2

 

75 = 5 + 2n - 2

 

75 = 2n +3

 

72 = 2n

 

n = 36

 

 

Sum of first n terms = (n/2)(a₁ + a_n)

 

S = (36/2)(5 + 75)

S = 18(80)

S_n =    1440

 

 

b)

 

a₁ = x+1

d = 2x+1 - (x+1) = 2x + 1 - x -1 = x

 

a_n = a₁ + (n-1)d

a_n = x+1 + (n-1)x   = x+1 + xn - x = xn + 1

21x + 1 = xn + 1

21x = xn

 

n = 21

 

 

S_n = (21/2)(21x+1 + (x+1))

    = (21/2)(22x + 2)

    = (21)(11x + 1)

 

S_n = 231x + 21

 

 

Check with x = 1:

 

Sn = 231(1) + 21 = 252

 

a₂₁ = (21/2) (1+1 + 21(1) + 1)

     = (21/2) (24)

     = 21 * 12 = 252