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.
= n^{2}^{ } (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 = 1^{2} + 2^{2} + 3^{2} + ............. + 40^{2}
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
11^{3} + 12 ^{3} + ............... +k^{3} where k = 50
Solution :
By applying the value of k, we get
(11^{3} + 12^{3} + ..........+50^{3})
Step 1 :
Find the sum of the series from 1 upto the last term 50.
(1^{3} + 2^{3} + ..........+ 50^{3})
Sum of cubes = [n(n + 1)/2]^{2}
n = 50
= [50(50 + 1)/2]^{2}
= [25 ⋅ 51]^{2}
= 1275^{2}
= 1625625 ----(1)
Step 2 :
Find the sum of the series from 1 upto the preceding term of the first term in the given series.
(1^{3} + 2^{3} + ..........+ 10^{3})
= [10(10 + 1)/2]^{2}
= (5 ⋅ 11)^{2}
= 55^{2}
= 3025 ----(2)
Step 3 :
(1) = (2)
Required result = 1625625 - 3025
= 1622600
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