**Series** can be defined as the sum of all the numbers of the given sequence. The sequences are finite as well as infinite. In the same way, the series can also be finite or infinite. For example, consider a sequence as **1, 3, 5, 7, …** Then the series of these terms will be **1 + 3 + 5 + 7 + …. **. The series special in some way or the other is called a special series. The following are the **three types of special series.**

**1 + 2 + 3 +… + n (sum of first n natural numbers)****1**^{2}+ 2^{2}+ 3^{2}+… + n^{2}(sum of squares of the first n natural numbers)**1**^{3}+ 2^{3}+ 3^{3}+… + n^{3 }(sum of cubes of the first n natural numbers)

In this article, we will see how to get the formula for all these series.

## Special Series 1: S**um of first n natural numbers**

The Result of this series is given below:

1+ 2 + 3 + 4 + …. + n = n (n + 1) / 2

**Proof:**

Let S

_{n}= 1 + 2 + 3 + 4 + … + nWe can see that this is an Arithmetic Progression with the first term (a) = 1 and common difference (d) =1 and there are n term

So, Sum of n terms = n/2 (2 x a + (n – 1) x d)

Putting the values for this series we will get

S

_{n }= n/2(2 x 1 + (n – 1) x 1)S

_{n}= n/2(2 + n – 1)S

_{n }= n(n + 1)/2

Hence Proved.

**Example **

**Question. Find the sum of the following series 3 + 4 + 5 —- + 25?**

**Solution:**

Let S

_{n}= 3+ 4 + 5 — + 25Now we can also write it like this

S

_{n}+ 1 + 2 = 1 + 2 + 3 + 4 —- + 25Clearly now it is the sum of first 25 natural number we can be written like this

S

_{n}+ 1+ 2 = 25 (25 + 1) / 2S

_{n}= 325 – 1 – 2S

_{n}= 322

## Special Series 2: S**um of squares of the first n natural numbers**

The Result of this series is given below:

1^{2}+ 2^{2}+ 3^{2}+… + n^{2 }= n(n + 1) (2n + 1)/6

**Proof:**

Let S

_{n }=_{ }1^{2}+ 2^{2}+ 3^{2}+… + n^{2}—eq 1We know that, k

^{3}– (k – 1)^{3 }= 3k^{2}– 3k + 1— eq 2We know that, (a – b)

^{3}= a^{3}– b^{3}– 3a^{2}b + 3ab^{2}So, k

^{3}– (k – 1)^{3}= k

^{3}– k^{3}+1 + 3k^{2}– 3k= 3k

^{2}– 3k +1Putting k = 1, 2…, n successively in

eq 2,we obtain1

^{3}– 0^{3}= 3(1)^{2}– 3(1) + 12

^{3}– 1^{3}= 3(2)^{2}– 3(2) + 13

^{3}– 2^{3}= 3(3)^{2}– 3(3) + 1…………………………………

…………………………………

………………………………..

n

^{3}– (n – 1)^{3}= 3(n)^{2 }– 3(n) + 1Adding both sides of all above equations, we get

n

^{3}– 0^{3}= 3 (1^{2}+ 2^{2 }+ 3^{2}+ … + n^{2}) – 3 (1 + 2 + 3 + … + n) + nWe can write this like:

n

^{3 }= 3 ∑(k^{2}) – 3∑(k) +n, where 1 ≤ k ≤ n— eq(3)We know that,

∑(k) (where 1≤k≤n ) = 1 + 2 + 3 + 4 — n = n(n + 1)/2 —eq(4)and

eq 1can also be written like thisS

_{n }= ∑(k^{2}), where 1 ≤ k ≤ n— eq(1)Now, putting these values in

eq 3n

^{3 }= 3S_{n}– 3(n)(n + 1)/ 2 + nn

^{3}+ 3 (n) (n + 1)/2 – n = 3S_{n}(2n

^{3 }+ 3n^{2}+ 3n – 2n)/2 = 3S_{n}(2n

^{3 }+ 3n^{2 }+ n)/6 = S_{n}n(2n

^{2 }+ 3n + 1)/6 = S_{n}n(2n

^{2 }+ n + 2n + 1)/6 = S_{n}n(n(2n + 1) + 1(2n + 1))/6 = S

_{n}n(n + 1)(2n + 1)/6 = S

_{n}S

_{n }= n(n + 1)(2n + 1)/6

Hence proved.

**Examples **

**Question 1. Find the sum of the n terms of the series whose nth terms is n ^{2 }+ n + 1?**

**Solution: **

Given that ,

a

_{n }= n^{2 }+ n + 1Thus, the sum to n terms is given by

S

_{n }= ∑ak(where 1 ≤ k ≤ n )= ∑ k^{2}+ ∑ 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

= ((n

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

^{3 }+ 2n^{2 }+ n^{2}+ n + 3n^{2}+ 9n)/6= (2n

^{3}+ 6n^{2 }+ 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 (i

^{2}+ i)= (1/2)(Σ i

^{2}+ Σ i)We know Σ i

^{2}= n (n + 1) (2n + 1) / 6 andΣ 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

## Special Series 3: S**um of cubes of the first n natural numbers**

The Result of this series is given below:

1^{3}+ 2^{3}+ 3^{3}+ … + n^{3}= (n (n + 1)/2)^{2}

**Proof:**

Let S

_{n}= 1^{3}+ 2^{3}+ 3^{3}+… + n^{3}—eq 1We know that, (k + 1)

^{4}– (k)^{4}= 4k^{3}+ 6k^{2}+ 4k + 1— eq 2We know that, (a+b)

^{4}= (a^{2}+b^{2}+2ab)^{2}= a

^{4}+ b^{4}+ 6a^{2}b^{2}+ 4a^{3}b + 4ab^{3}So, (k + 1)

^{4}– (k)^{4}= k

^{4}+ 1 + 6k^{2}+ 4k^{3}+ 4k- k^{4}= 4k

^{3}+6k^{2}+ 4k +1Putting k = 1, 2…, n successively in

eq 2, we obtain(1 + 1)

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

^{4}– 2^{4}= 4(2)^{3}+ 6(2)^{2 }+ 4(2) + 1…………………………………

…………………………………

………………………………..

(n + 1)

^{4}– (n)^{4 }= 4(n)^{3}+ 6n^{2}+ 4n + 1Adding both sides of all the above equations, we get

(n + 1)

^{4}– 1^{4}= 4 (1^{3}+ 2^{3}+ 3^{3}+ … + n^{3}) + 6(1^{2 }+ 2^{2}+ 3^{2}+ 4^{2}+ 5^{2}) + 4 (1 + 2 + 3 + … + n) + nWe can write this like:

(n + 1)

^{4 }– 1^{4 }= 4 ∑ (k^{3}) + 6∑(k^{2}) + 4∑(k) + n where 1 ≤ k ≤ n— eq(3)We know that ,

∑(k) (where 1 ≤ k ≤ n ) = 1 + 2 + 3 + 4 — n = n (n + 1)/2

—eq(4)∑(k

^{2}) (where 1 ≤ k ≤ n ) = 1^{2 }+ 2^{2 }+ 3^{2 }+ 4^{2}— n^{2}= n (n + 1) (2n + 1)/6—eq(5)and

eq 1can also be written like thisS

_{n }= ∑(k^{3}) , where 1 ≤ k ≤ n — eq(1)Now, putting these values in

eq 3(n + 1)

^{4}-1^{4 }= 4S_{n}+ 6(n) (n + 1) (2n + 1)/6 + 4 (n) (n + 1)/2 + nn

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

^{4}+ 6n^{2}+ 4n^{3}+ 4n – 2n^{3}– 3n^{2}– n – 2n^{2}– 2n – n = 4S_{n}n

^{4}+ n^{2}+ 2n^{3}= 4S_{n}n

^{2 }(n^{2 }+ 1 + 2n) = 4S_{n}n

^{2 }(n + 1)^{2}= 4S_{n}S

_{n }= (n(n + 1)/2)^{2}

Hence proved.

**Example **

**Question. Find the value of the following fraction (1 ^{3} + 2^{3} + 3^{3} —- + 9^{3}) / (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, (1

^{3}+ 2^{3}+ 3^{3}—-+ n^{3}) / (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