When you go shopping for a shirt or t-shirts, you do scan the colour shades of bundles arranged in racks. If they are arranged in such a manner that you could get the shirt you like as early as possible then you think of the shop or market as good. What if the clothes are not arranged in series? You will definitely not visit the place again. Let us now try to understand the topic special series in detail.

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**What is a Series?**

*(Source: wikihow)*

We can define the series 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 I 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 + …

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Learn Sequences and Series here.

## What are the Special Series?

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)

Let’s now look at the sum to n terms of special series:

### Sum of First n Natural Numbers (1 + 2 + 3 +… + n)

The above series is an AP. Here a = 1, d =1 because the difference between these terms is 1 and also 1st term is 1.

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

S_{n }= \( \frac{n}{2} \) [2× 1 + ( n-1 )1]

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

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

So above is the formula of the seies.

### Sum of Squares of the First n Natural Numbers (1^{2} + 2^{2} + 3^{2} +… + n^{2})

This series is neither in GP nor in AP. So let’s first convert this in GP or AP. Here we use the formula,

k^{3 }– (k-1)^{3}= k^{3 }– (k-1)^{3} = 3k^{2 }– 3k +1

Let us take the value of k as k = 1

1^{3} – 0^{3} = 3 (1)^{2} – 3 (1) + 1

Now take k =2

2^{3} – 1^{3} = 3 (2)^{2} – 3 (2) + 1

Now k = 3

3^{3} – 2^{3} = 3(3)^{2} – 3 (3) + 1

Similarly if we take k =n, we get,

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

Now add both the sides, the equation we get is,

n^{3} – 0^{3} = 3 (1^{2} + 2^{2} + 3^{2} + … + n^{2}) – 3 (1 + 2 + 3 + … + n) + n

n^{3 }= 3 (S_{n}) – 3 × \( \frac{n(n+1)}{2} \) + n

3 S_{n}= n³ + 3 \( \frac{n²+ n)}{2} \) – n

S_{n }= \( \frac{n(n+ 1)(2n+1)}{6} \)

### Sum of Cubes of the First n Natural Numbers (1^{3} + 2^{3} + 3^{3} +… + n^{3})

Here we use the formula,

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

Let us take the value of k as k = 1

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

Now take the value of k = 2

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

Again take k = 3

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

Similarly if we take k =n

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

Now adding both the sides, we get

(n + 1)^{4} – 1^{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

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

= n^{4}+ 4n³ + 6n² + 4n- n (2n² + 3n +1 )-2n(n+1)-n

= n^{4}+ 2n³ + n²

= n² (n+1)²

S_{n }= \( \frac{n²(n+1)²}{4} \)

= \( \frac{n(n+1)²}{4} \)

Hence we get the formula for all the three special cases.

## Solved Examples for You

Question: 2, 3, 6, n, 42… Find the value of n in the above series.

- 9
- 12
- 15
- 18
- 21

Solution: C is the correct option. Given series is 2, 3, 6, n, 42… Here we can decode the series by multiplying the previous number by 3 and then by subtracting 3 from it. As we can see 2 being the first numbers of the series.

2 × 3 = 6

6 – 3 = 3 (3 is the second number)

3 × 3 -3 = 6 ( 6 is the third number)

n will be 6 × 3 – 3 = 18-3 = 15

Hence, n is 15

Question: Consider the following statements:

(1) The sum of cubes of first 2o natural numbers is 44400

(2) The sum of squares of first 20 natural numbers is 2870

Which of the above statements is/are correct?

- Only 1st
- Only 2nd

Solution: B is the correct option. The sum of the cubes of first n consecutive natural numbers is given by \( \frac{n(n+1)}{2} \)²

Putting n = 20, we have sum = \( \frac{20(20+1)}{2} \)² = 44100

The sum of the squares of first n consecutive natural numbers is given by \( \frac{n(n+ 1)(2n+1)}{6} \)

Putting n = 20, we have the sum = \( \frac{20(20+ 1)(2×20+1)}{6} \) = 2870

Ques. What are the various types of series?

Ans. There are various types of series under the mathematics subject. Some of them are given below:

Harmonic Series: This series is one of the examples of the divergent series.

Geometric Series: In the geometric series the ratio of each 2 consecutive terms is a continuous function of the summary index.

P Series: P series is the series where the mutual exponent P is a positive and real constant number.

Ques. What is the limit of a series?

Ans. Limit of a Series defines that ‘the number s is known as the sum of the series. If the series not converges, the series will be said to be divergent, and we would say that the series is diverging’.

Ques. What is a Series formula?

Ans. A series has a similar and constant difference between its terms. Such as; 3 + 7 + 11 + 15 +… and so on. We can find the sum of the arithmetic series through the process of multiplying the number of times with the average of the first and the last terms.

Ques. What is the sum of a series?

Ans. The n-th incomplete sum of any series is basically the sum of all the first n terms. The order of the partial sums of a series occasionally tends to a real limit.