HOW TO FIND THE SUM OF SPECIAL SERIES

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

Here we study some common special series like

(i)  Sum of first "n" natural numbers.

  =  n(n + 1)/2

(ii)  Sum of first "n" odd natural numbers. 

 =  n2  (or)  [(l + 1)/2]2

(iii)  Sum of squares of "n" natural numbers :

  =  n(n + 1) (2n + 1) / 6

(iv)  Sum of cubes of "n" natural numbers :

  =  [n(n + 1) / 2]2

Note :

We can use the formulas given above only if the given series starts from 1.

If it doesn't start from 1, we have to follow the steps given below to find the required sum.

Step 1 :

Find the sum from 1 upto the last term we have in the given series.

Step 2 :

Find the sum from 1 upto the preceding term of the first term what we have in the given series.

Step 3 :

By subtracting the results from (1) and (2), we will get the required sum. That is,

Required sum  =  Result from step 1 - Result from step (2)

Example 1 :

Find the sum of first 75 natural numbers.

Solution :

Sum of first 75 natural numbers

1 + 2 + 3 + 4 + ................. + 75

Sum of natural numbers  =  n(n + 1)/2

Here "n" stands for total number of terms.

  =  75 (75 + 1)/2

  =  (75  76) / 2

  =  75  38

  =  2850

Therefore the sum of first 75 natural number is 2850.

Example 2 :

Find the sum of

15 + 16 + 17 + .............. + 80

Solution :

Since the given series doesn't start from 1, we are decomposing the given series into two parts.

Step 1 :

Find the sum of the series from 1 upto the last term 80.

(1 + 2 + 3 +........+ 80)

Sum of natural numbers  =  n(n + 1) / 2

n  =  80

  =  80(80 + 1) / 2

  =  40 ⋅ 81

  =  3240  -----(1)

Step 2 :

Find the sum of the series from 1 upto the preceding term of the first term in the given series.

(1 + 2 + 3 +........+ 14)

Sum of natural numbers  =  n(n + 1) / 2

n  =  14

  =  14(14 + 1) / 2

  =  7 ⋅ 15

  =  105  -----(2)

Step 3 :

(1)  =  (2)

Required sum  =  3240 - 105

  = 3135 

Example 3 :

Find the sum of

1 + 4 + 9 + ............. + 1600

Solution :

By observing the given series, we find that every term can be represented in the form of square.

1 + 4 + 9 + ............. + 1600  =  12 + 22 + 32 + ............. + 402

Sum of squares  =  n(n+ 1) (2n + 1) / 6

n  =  40

  =  40 (40 + 1) (2⋅40 + 1) / 6 

  =  (40 ⋅ 41 ⋅ 81) / 6

  =  22140

Example 4 :

Find the sum of the following series

113 + 12 3 + ............... +k3 where k = 50

Solution :

By applying the value of k, we get

(113 + 123 + ..........+503)

Step 1 :

Find the sum of the series from 1 upto the last term 50.

(13 + 23 + ..........+ 503)

Sum of cubes  =  [n(n + 1)/2]2

n  =  50

  =  [50(50 + 1)/2]2

  =  [25 ⋅ 51]2

  =  12752

  =  1625625  ----(1)

Step 2 :

Find the sum of the series from 1 upto the preceding term of the first term in the given series.

(13 + 23 + ..........+ 103)

  =  [10(10 + 1)/2]2

  =  (5 ⋅ 11)2

  =  552

  =  3025  ----(2)

Step 3 :

(1)  =  (2)

Required result  = 1625625 - 3025

  =  1622600

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Permutation and Combination Problems

    Nov 27, 22 08:59 PM

    Permutation and Combination Problems

    Read More

  2. Combination Problems With Solutions

    Nov 27, 22 08:56 PM

    Combination Problems With Solutions

    Read More

  3. Like and Unlike Fractions Definition

    Nov 26, 22 08:22 PM

    Like and Unlike Fractions Definition - Concept - Examples with step by step explanation

    Read More