Sum of N Terms of Special Series

Series is defined as the sum of all given sequence numbers. The series can be finite or infinite depending upon the sequence of the numbers. For the given sequence as  2, 4, 6, … the series for this sequence will be 2+4+6+…

There are two types of progression in series.

• Arithmetic Progression
• Geometric Progression 

The special series is defined as a unique series in one way or another. There are three types of special series.

• 1+2+3+4+…+n (Sum of first n terms) 
• 12+22+32+42+…+n2 (Sum of squares of first n terms)
• 13+23+33+43+…+n3(Sum of cubes of first n terms)

1+2+3+4+…+n (Sum of first n terms)

Here, we need to find the sum of the first n natural numbers. As we can see, this type of series is an arithmetic progression series. Therefore, a=1 and d=1

The formula for the sum of A.P is 

Sn =(n/2)[2a +(n-1)d)

Sn= (n/2)[2*1 +(n-1)1]

Sn= (n/2)[2+(n-1)]

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

The above formula is for the special series of the sum of first n natural numbers.

12+22+32+42+…+n2 (Sum of squares of n terms)

The following steps are used to calculate the sum of squares of n terms. We know that this series cannot be arithmetic or geometric. So, we need to convert the series into the ones mentioned above to calculate the formula.

Here, we use this formula.

k3-(k-1)3=3k2+3k+1

We will keep the value of k as 1

13-03=3(1)2-3(1)+1

23-13=3(2)2-3(2)+1

33-23=3(3)2-3(3)+1

Similarly, if we keep k as n, then.

n3-(n-1)3=3(n)2-3(n)+1

Now when we add both sides of the equations we get,

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

n3 = 3Sn- 3*[n(n+1)]/2 +n

3 Sn= n3 + 3 [(n2+n)/2 ] – n

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

The above is the formula calculated for the sum of squares of first n natural numbers. 

13+23+33+43+…+n3(Sum of cubes of n terms)

The following steps will be used to calculate the formula for this type of special series.

We will apply this formula here,

(k+1)4 – k4 =4 k3 +6 k2 + 4k + 1

We keep the value of k as 1

24 – 14 =4 (13) +6 (12)+ 4 + 1

Here, we will keep the value of k as 2

(3)4 – 24 =4 (23 )+6 (22) + 4*2 + 1

Here, keep the value of k as 3

(4)4 – 34 =4 (33 )+6 (32) + 4*3 + 1

Similarly, if we keep k as n

(n+1)4 – n4 =4 n3 +6 n2 + 4n + 1

Now after we add both the sides we will get the following ,

(n+1)4- 14=4(13+23+33+43+…+n3)+6(12+22+32+42+…+n2)+4(1+2+3+4+…+n)+n

(n+1)4= 4 Sn+6*[n(n+1)(2n+1)]/6 + 4*(n/2)[n+1]+n

= n4 +4n3 + 6n2+ 4n -n(2n2+3n+1)-2n(n+1)-n

=n4+2n3+n2

=n2(n+1)2

Sn=[ n2(n+1)2]/4

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

The above is the result for calculating the sum of cubes of n natural terms.

Solved Examples

Example 1: Find the sum of the first n terms of the series 1.3+ 3.5 +5.7 + ….

Solution:

Let Sn= 1.3 + 3.5 + 5.7 + …

Here is the nth term of the series

Tn={nth term of 1, 3, 5, …} *{nth term of 3, 5, 7, …

    = (2n-1) (2 n +1) = 4 n2+1

Sn= Σ tn

    = 4 Σ n2 -Σ(1)

    = 4 Σ n2 -Σ(1)

    = {4[n(n+1)(n+2)(2n+1)]/6}-n

    = {2[n(n+1)(2n+1)]-3n}/3

    = (n/3)[2(2n2+ 3n +1)-3]

    = (n/3)[4n2+6n -1]

Example 2: Find the value of the following fraction (13+23+33+43+…+93<?sup>)/(1+2+3+4+…+9)

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

                Sum of cube of first n natural numbers: [n(n+1)]/2]2

So (13+23+33+43+…+n3)/(1+2+3+4+…+n) 

    = [n(n+1)]/2]2n(n+1)/2

    = n(n+1)/2

Now as we can see that the value n is 9

= 9 (9 +1)/2

= 9 * 5

= 45

Conclusion

Sequences and series are important topics for our studies and our daily lives. I hope you can understand the special series with examples and explanations. We hope that you have a better understanding of the topic. This topic gives an overview of solving problems related to the sum of n terms of the special series.