SPECIAL SERIES EXAMPLES

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 +......+ k³  =  [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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