HOW TO FIND SUM OF SPECIAL SERIES

How to Find Sum of Special Series ?

There are some series whose sum can be expressed by explicit formulae. Such series are called special series.

Formula Used in Finding Sum of Special Series

Here we study some common special series like

(i) Sum of first ‘n’ natural numbers

1 + 2 + 3 + ............ + n  =  n(n + 1)/2

(ii) Sum of first ‘n’ odd natural numbers.

1 + 3 + 5 + ............ + (2n-1)  =  n2

(iii) Sum of squares of first ‘n’ natural numbers.

12 + 22 + 32 + ............ + n2  =  n(n + 1)(2n + 1)/6

(iv) Sum of cubes of first ‘n’ natural numbers.

13 + 23 + 33 + ............ + n3  =  [n(n + 1)/2]2

How to Find Sum of Special Series - Questions

Question 1 :

Find the sum of the following series

(i) 1 + 2 + 3 +............+ 60

Solution :

Sum of natural numbers 

1 + 2 + 3 +...............+ n  =  n(n + 1)/2

n = 60

1 + 2 + 3 +............+ 60  =  60(60+ 1 )/2

 =  30(61)

=  1830

(ii) 3 + 6 + 9 + ............+ 96

Solution :

  =  3 + 6 + 9 +...........+ 96

  =  3(1 + 2 + 3 + ............+ 32)

  =  3 (32(32+1)/2)

  =  3 (16) (33)

  =  1584

(iii) 51+ 52 + 53 +............+ 92

Solution :

  51+ 52 + 53 +............+ 92

  =  (1 + 2 + 3 +..........+ 92) - (1 + 2 + 3.........+ 50)

 n = 92 & n = 50 

=  [92(92 + 1)/2] - [50(50 + 1)/2]

=  46(93) - 25(51)

  =  4278 - 1275

=  3003

(iv) 1 + 4 + 9 + 16 +...............+ 225

Solution :

1 + 4 + 9 + 16 +...............+ 225

  =  12 + 22 + 32 + 42 +...............+ 152

Sum of squares  =  n(n + 1)(2n + 1)/6

  =  15(15 + 1) (2(15) + 1)/6

  =  15(16)(31)/6

  =  5(8) (31)

  =  1240

(v) 62 + 72 + 82 +........+212

Solution :

62 + 72 + 82 +........+212  

  =  (12 + 22 +........+ 212) - (12 + 22 + .......+ 52)

Sum of squares  =  n(n + 1)(2n + 1)/6

  =  [21(21 + 1) (2(21) + 1)/6] - [5(5 + 1) (2(5) + 1)/6]

  =  [21(22) (43)/6] - [5(6) (11)/6]

  =  [21(22) (43)/6] - [5(6) (11)/6]

  =  3311 - 55

  =  3256

(vi) 103 + 113 + 123 +.........+ 203

Solution :

103 + 113 + 123 +.........+ 203

  =  (13 + 23 + 33 +.........+ 203) - (13 + 23 + 33 +.........+ 93)

Sum of cubes  =  [n(n + 1)/2]2

  =  [20(20+1)/2]2 - [9(9+1)/2]2

  =  2102 - 452

  =  (210 + 45) (210 - 45)

  =  255 (165)

  =  42075

(vii) 1 + 3 + 5 +.............+ 71

Solution :

Sum of odd numbers  =  ((l+1)/2)2

  =  [(71 + 1)/2]2

  =  [72/2]2

 =  362

 =  1296

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Sum of Special Series". 

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