SPECIAL SERIES EXAMPLES

Special series are the series which are special in some way. 

Some of the special series are : 

(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

Example 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

Example 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

Example 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.

Example 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

Example 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

Example 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

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