Sum to n Terms of Special Series for JEE | Solved Examples, FAQs, Practice Problems

📢 Sum to n Terms of Special Series – Class 11 Mathematics 📚

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The topic Sum to n Terms of Special Series is an important part of Class 11 Mathematics, covering sequences like sum of natural numbers, sum of squares, sum of cubes, and more. Understanding these series helps students excel in CBSE Board Exams, JEE Mains, and Advanced, as they frequently appear in algebra, calculus, and problem-solving.

What are Special Series?

Special series in mathematics refers to specific sequences of numbers or terms that follow a particular pattern or rule and are widely studied for their unique properties and applications. These series include well-known examples like the arithmetic series, geometric series, harmonic series, and Fibonacci series.

Sum of n Terms of Special Series

Some special series are given below:

(i) 1 + 2 + 3 +…+ n (sum of first n natural numbers)

(ii) 1² + 2² + 3² +…+ n² (sum of squares of the first n natural numbers)

(iii) 1³ + 2³ + 3³ +…+ n³ (sum of cubes of the first n natural numbers)

In this article, we will read about special series in maths and how to derive the formula for all these series. We will also see solved examples and practice problems on special series.

Sum of First n Natural Numbers

Natural numbers are: 1, 2, 3, 4,….

Sum of these natural numbers can be written as: 1 + 2 + 3 + 4 +….

This is an AP with first term 1 and common difference 1.

i.e. a = 1 and d = 2 – 1 = 1

Sum of first n terms of an AP = n/2 [2a + (n – 1)d]

Now,

S_n = 1 + 2 + 3 + 4 + ….. + n

S_n = n/2 [2a + (n – 1)d]

Substituting a = 1 and d = 1,

S_n = (n/2) [2(1) + (n – 1)(1)]

= (n/2) [2 + n – 1]

= n(n + 1)/2

Therefore, the sum of first n natural numbers = n(n + 1)/2

Example on Sum of first n natural numbers

Question : Find the sum of the following series 4 + 5 +……….. + 25?

Solution:

Let S_n = 4 + 5 +….. + 25

Now we can also write it like this 

S_n + 1 + 2 +3 = 1 + 2 + 3 + 4 —- + 25

Clearly now it is the sum of first 25 natural number we can be written like this 

S_n + 1+ 2 +3 = 25 (25 + 1) / 2

S_n = 325 – 1 – 2 – 3

S_n = 319

Sum of Squares of The First n Natural Numbers

The squares of natural numbers are: 1², 2², 3², 4²,…

Or

1, 4, 9, 16, ….

We can express the sum of n terms as: 1² + 2² + 3² +…+ n²

This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.

Let’s find the sum of this series by considering an expression given below:

k³ – (k – 1)³ = 3k² – 3k + 1

Substituting k = 1,

1³ – (1 – 1)³ = 3(1)² – 3(1) + 1

1³ – 0³ = 3(1)² – 3(1) + 1….(i)

Substituting k = 2,

2³ – (2 – 1)³ = 3(2)² – 3(2) + 1

2³ – 1³ = 3(2)² – 3(2) + 1….(ii)

Substituting k = 3,

3³ – (3 – 1)³ = 3(3)² – 3(3) + 1

3³ – 2³ = 3(3)² – 3(3) + 1….(iii)

Substituting k = 4,

4³ – (4 – 1)³ = 3(4)² – 3(4) + 1

4³ – 3³ = 3(4)² – 3(4) + 1….(iv)

…….

Substituting k = n,

n³ – (n – 1)³ = 3(n)² – 3(n) + 1

Now, adding both sides of these equations together, we get;

1³ – 0³ + 2³ – 1³ + 3³ – 2³ + … + n³ – (n – 1)³ = 3(1² + 2² + 3² + 4² + … + n²) – 3(1 + 2 + 3 + 4 + … + n) + n(1)

n³ – 0³ = 3(1² + 2²+ 3² + 4² + … + n²) – 3(1 + 2 + 3 + 4 + … + n) + n

\(\begin{array}{l}n^{3}=3\sum_{k=1}^{n}k^{2}-3\sum_{k=1}^{n}k+n\end{array} \)

Here,

\(\begin{array}{l}\sum_{k=1}^{n}k\end{array} \)

represents the sum of first n natural numbers and is equal to n(n + 1)/2.

So,

\(\begin{array}{l}n^{3}=3\sum_{k=1}^{n}k^{2}-3[\frac{n(n+1)}{2}]+n\end{array} \)

Rearranging the terms,

\(\begin{array}{l}\sum_{k=1}^{n}k^{2}=\frac{1}{3}[n^{3}+3[\frac{n(n+1)}{2}]-n]\end{array} \)

\(\begin{array}{l}\sum_{k=1}^{n}k^{2}=\frac{1}{6}[2n^{3}+3n^2+3n-2n]\end{array} \)

= (1/6) (2n³ + 3n² + n)

= (1/6) [n(2n² + 3n + 1)]

= (1/6)[n(n + 1)(2n + 1)]

Therefore, the sum of squares of first n natural numbers = [n(n + 1)(2n + 1)]/6

Examples on Sum of squares of the first n natural numbers

Question 1. Find the sum of the n terms of the series whose nth terms is n² + n + 1?

Solution: 

Given that , 

a_n = n² + n + 1

Thus, the sum to n terms is given by

S_n = ∑a_k (where 1 ≤ k ≤ n ) = ∑ k² +  ∑ k +  ∑1  (where 1 ≤ k ≤ n)

= n(n + 1) (2n + 1)/6 + n (n + 1)/2 + n

= (n(n + 1) (2n + 1) + 3n(n + 1) + 6n)/6

= ((n²+ n) (2n + 1) + 3n² + 3n + 6n)/6

= (2n³ + 2n² + n² + n + 3n² + 9n)/6

= (2n³ + 6n² + 10n)/6

Question 2. Find the sum of the following series up to n terms 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 3 + 4 +  ——?

Solution: 

If we observe the  series carefully we can write it like this 

S_n =(1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ——

We can say that we have to find sum of the sum of  first n natural number.

So we can write S_n=  Σ((i(i + 1))/2), where 1 ≤ i ≤ n

= (1/2)Σ (i(i + 1))

= (1/2)Σ (i² + i)

= (1/2)(Σ i² + Σ i)

We know Σ i² = n (n + 1) (2n + 1) / 6 and  

Σ i = n (n + 1) / 2.

Substituting the value, we get,

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

        = n(n + 1)/2 [(2n + 1)/6 + 1/2]

        = n(n + 1)(n + 2) / 6

Sum of Cubes of The First n Natural Numbers

The squares of natural numbers are: 1³, 2³, 3³, 4³,…

Or

1, 8, 27, 64,….

We can express the sum of n terms as: 1³ + 2³ + 3³ +…+ n³

This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.

Let’s find the sum of this series by considering an expression given below:

(k + 1)⁴ – k⁴ = 4k³ + 6k² + 4k + 1

Substituting k = 1, 2, 3, …, n

2⁴ – 1⁴ = 4(1)³ + 6(1)² + 4(1) + 1

3⁴ – 2⁴ = 4(2)³ + 6(2)² + 4(2) + 1

4⁴ – 3⁴ = 4(3)³ + 6(3)² + 4(3) + 1

…..

(n – 1)⁴ – (n – 2)⁴ = 4(n – 2)³ + 6(n – 2)² + 4(n – 2) + 1

n⁴ – (n – 1)⁴ = 4(n – 1)³ + 6(n – 1)² + 4(n – 1) + 1

(n + 1)⁴ – n⁴ = 4n³ + 6n² + 4n + 1

By adding both sides of these equations, we get;

2⁴ – 1⁴ + 3⁴ – 2⁴ + 4⁴ – 3⁴ + …. + (n + 1)⁴ – n⁴ = 4(1)³ + 6(1)² + 4(1) + 1 + 4(2)³ + 6(2)² + 4(2) + 1 + 4(3)³ + 6(3)² + 4(3) + 1 + …. + 4n³ + 6n² + 4n + 1

(n + 1)⁴ – 1⁴ = 4(1³ + 2³ + 3³ +…+ n³) + 6(1² + 2² + 3² + …+ n²) + 4(1 + 2 + 3 +…+ n) + n

\(\begin{array}{l}n^4 + 4n^3 + 6n^2 + 4n + 1 – 1 =4\sum_{k=1}^{n}k^3 + 6\sum_{k=1}^{n}k^2 + 4\sum_{k=1}^{n}k+n\end{array} \)

We know that,

\(\begin{array}{l}\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\end{array} \)

And

\(\begin{array}{l}\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\end{array} \)

Thus,

\(\begin{array}{l}n^4 + 4n^3 + 6n^2 + 4n = 4\sum_{k=1}^{n}k^3 + 6[\frac{n(n+1)(2n+1)}{6}]+4[\frac{n(n+1)}{2}]+n\end{array} \)

By rearranging the terms,

\(\begin{array}{l}4\sum_{k=1}^{n}k^3 = n^4 + 4n^3 + 6n^2 + 4n – 6[\frac{n(n+1)(2n+1)}{6}]-4[\frac{n(n+1)}{2}]-n\\\sum_{k=1}^{n}k^3 = \frac{1}{4}[n^4 + 4n^3 + 6n^2 + 4n – n (2n^2 + 3n + 1) – 2n(n + 1) – n]\end{array} \)

= (1/4) [n⁴ + 4n³ + 6n² + 4n – 2n³ – 3n² – n – 2n² – 2n – n]

= (1/4) [n⁴ + 2n³ + n²]

= (1/4)[n²(n² + 2n + 1)]

= (1/4)[n²(n + 1)²]

Therefore, the sum of cubes of first n natural numbers = [n(n + 1)]²/4

Example on Sum of cubes of the first n natural numbers

Question. Find the value of the following fraction (1³ + 2³ + 3³ —- + 9³) / (1 + 2 + 3 —- + 9)?

Solution: 

Sum of first n natural number : n(n + 1)/2

Sum of cube of first n natural number : (n(n + 1)/2)²

So, (1³ + 2³ + 3³ —-+  n³) / (1+ 2+ 3 —- +n)

= ((n(n + 1)/2)²) / (n(n + 1)/2)

= n(n + 1)/2

Now as we can see that value of n is 9 in the question,

= 9 (9 + 1) / 2

= 9 x 5

= 45

Solved Example for Sum to n Terms of Special Series for JEE

Question:

Find 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 S_n be the sum of its n terms and an be its nth term. Then,

S_n =2 + 5 + 14 + 41 + … + a_n….(i)

And

S_n = 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

⇒ a_n = 2 + [3 + 9 + 27 +… + (n- 1) term]

Here, 3 + 9 + 27 + … is a geometric series.

So, a_n = 2 + [3(3^n-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

Sn = [(1 + 3)/2] + [(1 + 3²)/2] + [(1 + 3³)/2] + …. + (1 + 3ⁿ)/2

= (1/2) [(3 + 3² + 3³ + …. + 3ⁿ) + (1 + 1 + 1 + … + n)]

= (1/2) {[3(3ⁿ – 1)/(3 – 1)] + n}

= (1/2) [(3/2)(3ⁿ – 1) + n]

= [(3ⁿ⁺¹ – 3 + 2n)/4]

Therefore, the sum of the given series = (3ⁿ⁺¹ + 2n – 3)/4

Practice Problems on Special Series

1. Find the sum of the first 20 terms of the arithmetic series where the first term is 5 and the common difference is 3.

2. Calculate the sum of the first 8 terms of the geometric series where the first term is 2 and the common ratio is 3.

3. Find the sum of the first 5 terms of the harmonic series.

4. Find the 10th term in the Fibonacci series.

FAQs on Special Series

What is an arithmetic series and how do you find its sum?

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. To find the sum of the first nnn terms of an arithmetic series, you can use the formula: S_n = n²(2a + (n – 1)d)

How do you calculate the sum of a geometric series?

A geometric series is a sequence of numbers where each term is multiplied by a constant ratio to get the next term. The sum of the first nnn terms of a geometric series can be calculated using the formula:

[Tex]S_n = a \frac{r^n – 1}{r – 1}
[/Tex]

What is a harmonic series and where is it used?

A harmonic series is a sequence of numbers where each term is the reciprocal of an integer. The nnnth harmonic number is the sum of the reciprocals of the first nnn natural numbers:

Hⁿ = 1 + 1/2 + 1/3 + 1/4 ………..

What is the Fibonacci series and how is it generated?

The Fibonacci series is a sequence of numbers where each term is the sum of the two preceding ones, starting from 0 and 1. The first few terms of the Fibonacci series are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.

🔍 Applications of Special Series

📌 Mathematical Analysis – Used in calculus, algebra, and statistics
📌 Physics & Engineering – Motion equations, electrical circuits, mechanics
📌 Computer Science – Algorithm analysis, summation formulas in programming
📌 Economics & Finance – Growth models, financial forecasting

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