SUM OF SPECIAL SERIES EXAMPLES

Sum of Special Series Examples :

There are some series whose sum can be expressed by explicit formulae. Such series are called special series.

Here we are going to see some example problems on special series.

Formula Used in Finding Sum of Special Series

Here we study some common special series like

(i) Sum of first ‘n’ natural numbers

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

(ii) Sum of first ‘n’ odd natural numbers.

1 + 3 + 5 + ............ + (2n-1)  =  n2

(iii) Sum of squares of first ‘n’ natural numbers.

12 + 22 + 32 + ............ + n2  =  n(n + 1)(2n + 1)/6

(iv) Sum of cubes of first ‘n’ natural numbers.

13 + 23 + 33 + ............ + n3  =  [n(n + 1)/2]2

Sum of Special Series Examples - Questions

Question 1 :

If 1 + 2 + 3 + ...........+ k = 325 , then find 13 +23 + 33 +......+ k3.

Solution :

Given that :

1 + 2 + 3 + ...........+ k = 325

13 +23 + 33 +......+ k=  [k(k+1)/2]2

  =  (Sum of k natural numbers)2

  =  3252

  =  105625

Question 2 :

If 13 + 23 + 33 +............+ k3 = 44100 then find 1 + 2 + 3 +......+k .

Solution :

Given that 

 13 + 23 + 33 +............+ k3 = 44100

 [k(k + 1)/2]2  =  44100

 [k(k + 1)/2]  =  √44100

 [k(k + 1)/2]  =  210

1 + 2 + 3 + ......... + k  =  210

Question 3 :

How many terms of the series 13 + 23 + 33 +............  should be taken to get the sum 14400?

Solution :

Sn  =  14400

[n(n + 1)/2]2  =  14400

[n(n + 1)/2]  =  14400

[n(n + 1)/2]  =  120

n2 + n  =  240

n2 + n - 240  =  0

(n + 16)(n - 15)  =  0

n = -16, 15

Hence the required number of terms is 15.

Question 4 :

The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.

Solution :

12 + 22 + 32 +............ + n2  =  285

13 + 23 + 33 +............ + n3  =  2025 

By applying the value of n(n + 1)/2  =  45 

45[(2n + 1)/3]  =  285

(2n + 1)/3  =  285/45

(2n + 1)/3  =  57/9

(2n + 1)  =  57/3

6n + 3  =  57

6n  =  57 - 3  

6n  =  54

n = 9

Question 5 :

Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?

Solution :

Required area

102 + 112 + 122 + .............+ 242

  =  (12 + 22 + 32 + .........+ 242) - (12 + 22 + 32 + ..........+ 92)

  =  [24(24 + 1)(2(24) + 1)/6] -  [9(9 + 1)(2(9) + 1)/6]

  =  [24(25)(49)/6] -  [9(10)(19)/6]

  =  4900 - 285

  =  4615 cm2

Question 6 :

Find the sum of the series (23 − 1) + (43 −33) + (63 −53)+....... to (i) n terms (ii) 8 terms

Solution :

=  (23 − 1) + (43 − 33) + (63 − 53)+.......

First numbers are even and second numbers are odd.

(i)  n terms

General term of the given series  =  (2n)3 − (2n - 1)3

  =  8n3 - [(2n)3 - 3(2n)2 (1) + 3(2n)(1) - 13]

  =  8n3 - [8n3 - 12n2 + 6n - 1]

  =  8n3 - 8n3 + 12n2 - 6n + 1

  =  12n2 - 6n + 1

  =  [12n(n +1)(2n + 1)/6] - 6[n(n + 1)/2] + n

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

  =  2n(2n2 + n + 2n + 1) - 3n2 - 3n + n

  =  4n3 + 6n2 + 2n - 3n2 - 3n + n

  =  4n3 + 3n2

Hence the sum of n terms 4n3 + 3n2

(ii) 8 terms

  =  4n3 + 3n2

n = 8

  =  4(8)3 + 3(8)2

  =  2048 + 192

  =  2240

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Sum of Special Series". 

Apart from the stuff given in this section "How to Find Sum of Special Series"if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...