In this chapter, we will explore some special problems related to finding the sum of the series, including arithmetic series, geometric series, telescoping series, power series, etc. We will discuss various techniques and formulas to simplify and evaluate these series, along with examples to illustrate their applications. These types of problems are common in competitive exams and it is challenging to solve them within the stipulated time limit.

**Arithmetic Progression (A.P. Series)**

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed constant (common difference) to the preceding term.

A.P. Series= a, a+d, a+2d, a+3d, …….., a+(n-1)d

Here a is the first term and d is a common difference.

The n_{th} term of the series (T_{n})= a+(n-1)d

Let, the sum of the n terms of the series be S_{n}

S_{n}= a+(a+d)+(a+2d)+(a+3d)+ ……..+[a+(n-1)d]

Now write this series in reverse order

S_{n}= [a+(n-1)d]+[a+(n-2)d]…….+(a+3d)+(a+2d)+(a+d)+a

After adding both equations: (The sum of the series)

2S_{n}=[2a+(n-1)d]+[2a+(n-1)d]…………+[2a+(n-1)d]

2S_{n}=n×[2a+(n-1)d]

S_n=\frac{n}{2}[2a+(n-1)d] ……..(iii)

S_n=\frac{n}{2}[a+a+(n-1)d].

S_n=\frac{n}{2}[a+T_n].

S_n=\frac{a+T_n}{2}\times n …………(iv)

Since the sum of a series = average × number of terms

So the average of the AP series= \frac{a+T_n}{2}.

The average of the AP series=1/2×(first term+ last term)

**The sum of the first n Natural Numbers**

First n natural numbers are 1,2,3,4,5……,n

This is an A.P. series (Arithmetic Series) with a common difference of 1

First term (a)=1

Common difference (d)=1

The nth term of the series, T_{n}= n

The sum of the A.P. series (\Sigma{n})= \frac{n}{2}\times [2a+(n-1)\times d].

\Sigma{n}= \frac{n}{2}\times [2+(n-1)\times 1].

\Sigma{n}=\frac{n(n+1)}{2}…………..(i)

Note: \Sigma{n}=1+2+3+4+…….+n

\Sigma{n^2}=1^{2}+2^{2}+3^{2}+4^{2}+ …….+n^{2}

\Sigma{n^3}=1^{3}+2^{3}+3^{3}+4^{3}+ …….+n^{3}

**The sum of the squares of the first n Natural Numbers**

The squares of natural numbers: 1^{2}, 2^{2}, 3^{2}, 4^{2}, ……., n^{2}

The nth term of the series, T_{n}=n^{2}

n^{3}-(n-1)^{3}=n^{3}-{n^{3}-1^{3}-3×n×1(n-1)}

n^{3}-(n-1)^{3}= 3n^{2}-3n+1

Now, put n=1,2,3,….n (The sum of the series)

1^{3}-0^{3}= 3.1^{2}-3.1+1

2^{3}-1^{3}= 3.2^{2}-3.2+1

3^{3}-2^{3}= 3.3^{2}-3.3+1

…..

n^{3}-(n-1)^{3}= 3.n^{2}-3.n+1

After adding all the equations: (The sum of the series)

n^3- 0= 3.\Sigma{n^2}-3.\Sigma{n}+ (1+1+1+…..n times)

n^3= 3.\Sigma{n^2}-3.\frac{n(n+1)}{2}+n.

3.\Sigma{n^2}= n^3-n+3.\frac{n(n+1)}{2}.

= n.(n^2-1)+3.\frac{n(n+1)}{2}.

= n(n-1)(n+1)+3.\frac{n(n+1)}{2}.

= (n+1).[n^2-n+\frac{3}{2}n].

= (n+1).[n^2+\frac{1}{2}n].

=n(n+1).[n+\frac{1}{2}].

=\frac{1}{2}n(n+1)(2n+1).

\Sigma{n^2}=\frac{1}{6}n(n+1)(2n+1)………..(ii)

**The sum of the cubes of the first n Natural Numbers**

The cubes of natural numbers: 1^{3}, 2^{3}, 3^{3}, 4^{3}, ……., n^{3}

The nth term of the series, T_{n}=n^{3}

n^{4}-(n-1)^{4}={n^{2}-(n-1)^{2}}×{n^{2}+(n-1)^{2}}

⇒{n^{2}-(n^{2}-2n+1)}×{n^{2}+(n^{2}-2n+1)}

⇒(2n-1)×(2n^{2}-2n+1)

⇒4n^{3}-6n^{2}+4n-1

n^{4}-(n-1)^{4}= 4n^{3}-6n^{2}+4n-1

Now, put n=1,2,3,….n (The sum of the series)

1^{4}-0^{4}= 4.1^{3}-6.1^{2}+4.1-1

2^{4}-1^{4}= 4.2^{3}-6.2^{2}+4.2-1

3^{4}-2^{4}= 4.3^{3}-6.3^{2}+4.3-1

………..

n^{4}-(n-1)^{4}= 4n^{3}-6n^{2}+4n-1

After adding all the equations: (The sum of the series)

n^4-0= 4.\Sigma{n^3}-6.\Sigma{n^2}+4.\Sigma{n}-(1+1+1+…..n times)

4.\Sigma{n^3}= n^4 +6.\Sigma{n^2}-4.\Sigma{n}+n.

= n^4 +6.\frac{1}{6}n(n+1)(2n+1)-4.\frac{1}{2}n(n+1)+n.

= n^4 +n(n+1)(2n+1)-2n(n+1)+n.

= n.[n^3 +(2n^2+3n+1)-2n-2+1].

= n.[n^3 +2n^2+n].

= n^2.[n^2 +2n+1].

= n^2.(n+1)^2.

\Sigma{n^3}=[\frac{n(n+1)}{2}]^2 …………..(iii)

The nth root of a Number: Click Here