Sum of Some Special Series Formula: In mathematics, various types of series present themselves, including arithmetic series, geometric series, harmonic series, and more. In addition to these familiar series, we encounter special series for which we can ascertain the sum of their terms through distinct methods. Here, we will explore the process of deriving formulas to calculate the sum of these particular series up to 'n' terms. Our emphasis will be on three widely employed formulas, and we'll illustrate each one with solved examples.
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Summation of 'n' Terms in Special Series
Let's explore some special series:
(i) 1 + 2 + 3 +…+ n ( Sum of consecutive natural numbers from 1 to 'n.')
(ii) 1
2
+ 2
2
+ 3
2
+…+ n
2
( Sum of squares of natural numbers from 1 to 'n.)
(iii) 1
3
+ 2
3
+ 3
3
+…+ n
3
(Sum of cubes of natural numbers from 1 to 'n.')
Let us find the sum up to n terms of special series mentioned here, one by one.
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(i) Sum of First 'n' Natural Numbers
The set of natural numbers comprises 1, 2, 3, 4, …
The summation of these natural numbers can be expressed as: 1 + 2 + 3 + 4 + …
This forms an arithmetic progression (AP) with a first term '1' and a common difference '1'.
Therefore, we have 'a = 1' and 'd = 1'.
Utilizing the formula for the sum of the first 'n' terms of an AP:
S
n
= n/2 [2a + (n – 1)d]
Now, substituting 'a = 1' and 'd = 1', we obtain:
S
n
= (n/2) [2(1) + (n – 1)(1)] = (n/2) [2 + n – 1] = n(n + 1)/2
Hence, the sum of the first 'n' natural numbers is given by n(n + 1)/2.
(ii) Sum of Squares of the First 'n' Natural Numbers
The squares of natural numbers are: 1
2
, 2
2
, 3
2
, 4
2
, …
This series is neither an arithmetic progression (AP) nor a geometric progression (GP).
However, we can find its sum by considering the following expression:
k
3
– (k – 1)
3
= 3k
2
– 3k + 1
Substituting 'k = 1', we get:
1
3
– (1 – 1)
3
= 3(1)
2
– 3(1) + 1 = 1
Similarly, for 'k = 2', 'k = 3', and so on, we find corresponding expressions.
Now, adding both sides of these equations:
1
3
– 0
3
+ 2
3
– 1
3
+ 3
3
– 2
3
+ … + n
3
– (n – 1)
3
= 3(1
2
+ 2
2
+ 3
2
+ 4
2
+ … + n
2
) – 3(1 + 2 + 3 + 4 + … + n) + n
Using the sum of the first 'n' natural numbers formula (n(n + 1)/2), we simplify to:
n
3
– 0
3
= 3(1
2
+ 2
2
+ 3
2
+ 4
2
+ … + n
2
) – 3(n(n + 1)/2) + n
After rearranging terms, we arrive at the sum formula:
(1/6) [2n
3
+ 3n
2
+ n] = (1/6) [n(2n
2
+ 3n + 1)] = (1/6)[n(n + 1)(2n + 1)]
Therefore, the sum of the squares of the first 'n' natural numbers is [n(n + 1)(2n + 1)]/6.
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(iii) Sum of Cubes of the First 'n' Natural Numbers
The cubes of natural numbers are: 1
3
, 2
3
, 3
3
, 4
3
, …
Similar to the previous series, this is neither an AP nor a GP.
We can determine its sum using the following expression:
(k + 1)
4
– k
4
= 4k
3
+ 6k
2
+ 4k + 1
Substituting 'k' values from 1 to 'n', we obtain a series of equations.
Adding both sides of these equations, we arrive at:
2
4
– 1
4
+ 3
4
– 2
4
+ 4
4
– 3
4
+ …. + (n + 1)
4
– n
4
= 4(1
3
+ 2
3
+ 3
3
+ … + n
3
) + 6(1
2
+ 2
2
+ 3
2
+ … + n
2
) + 4(1 + 2 + 3 + … + n) + n
Utilizing the sum of squares of the first 'n' natural numbers formula ([n(n + 1)(2n + 1)]/6) and the sum of the first 'n' natural numbers formula (n(n + 1)/2), we simplify further to:
(1/4) [n
4
+ 4n
3
+ 6n
2
+ 4n – 2n
3
– 3n
2
– n – 2n
2
– 2n – n] = (1/4) [n
2
(n
2
+ 2n + 1)] = (1/4)[n
2
(n + 1)
2
]
Therefore, the sum of the cubes of the first 'n' natural numbers is [n(n + 1)]
2
/4.
Sum of Some Special Series Formula FAQs Solved Example
Question:
Determine the sum to n terms of the series: 2 + 5 + 14 + 41 +….
Solution:
2 + 5 + 14 + 41 +….
The difference between two consecutive terms of this series is: 3, 9, 27, ….
Let Sn be the sum of its n terms and an be its nth term. Then,
S
n
=2 + 5 + 14 + 41 + … + a
n
….(i)
And
Sn = 2 + 5+ 14 + 41 + … + a
n
– 1 + a
n
….(ii)
Subtracting equation (ii) from (i), we get
0 = 2 + [3 + 9 + 27 + … + (n – 1) term] – a
n
⇒ an = 2 + [3 + 9 + 27 +… + (n- 1) term]
Here, 3 + 9 + 27 + … is a geometric series.
So, an = 2 + [3(3n-1 – 1)/2]
= [4 + 3n – 3]/2
= (1 + 3n)/2
Now, we need to find the sum of the series whose general term is (1 + 3n)/2
S
n
= [(1 + 3)/2] + [(1 + 3
2
)/2] + [(1 + 3
3
)/2] + …. + (1 + 3
n
)/2
= (1/2) [(3 + 3
2
+ 3
3
+ …. + 3n) + (1 + 1 + 1 + … + n)]
= (1/2) {[3(3
n
– 1)/(3 – 1)] + n}
= (1/2) [(3/2)(3
n
– 1) + n]
= [(3
n+1
– 3 + 2n)/4]
Therefore, the sum of the given series = (3
n+1
+ 2n – 3)/4
