**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.

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) = n^{2}

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

1^{2} + 2^{2} + 3^{2} + ............ + n^{2} = n(n + 1)(2n + 1)/6

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

1^{3} + 2^{3} + 3^{3} + ............ + n^{3} = [n(n + 1)/2]^{2}

**Question 1 :**

If 1 + 2 + 3 + ...........+ k = 325 , then find 1^{3} +2^{3} + 3^{3} +......+ k^{3}.

**Solution :**

Given that :

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

1^{3} +2^{3} + 3^{3} +......+ k^{3 }= [k(k+1)/2]^{2}

= (Sum of k natural numbers)^{2}

= 325^{2}

= 105625

**Question 2 :**

If 1^{3} + 2^{3} + 3^{3} +............+ k^{3} = 44100 then find 1 + 2 + 3 +......+k .

**Solution :**

Given that

1^{3} + 2^{3} + 3^{3} +............+ k^{3} = 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 1^{3} + 2^{3} + 3^{3} +............ 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

n^{2} + n = 240

n^{2} + 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 :**

1^{2} + 2^{2} + 3^{2} +............ + n^{2} = 285

1^{3} + 2^{3} + 3^{3} +............ + n^{3} = 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

10^{2} + 11^{2} + 12^{2} + .............+ 24^{2}

= (1^{2} + 2^{2} + 3^{2} + .........+ 24^{2}) - (1^{2} + 2^{2} + 3^{2} + ..........+ 9^{2})

= [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 cm^{2}

**Question 6 :**

Find the sum of the series (2^{3} − 1) + (4^{3} −3^{3}) + (6^{3} −5^{3})+....... to (i) n terms (ii) 8 terms

**Solution :**

= (2^{3} − 1) + (4^{3} − 3^{3}) + (6^{3} − 5^{3})+.......

First numbers are even and second numbers are odd.

(i) n terms

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

= 8n^{3} - [(2n)^{3} - 3(2n)^{2} (1) + 3(2n)(1) - 1^{3}]

= 8n^{3} - [8n^{3} - 12n^{2} + 6n - 1]

= 8n^{3} - 8n^{3} + 12n^{2} - 6n + 1

= 12n^{2} - 6n + 1

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

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

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

= 4n^{3} + 6n^{2} + 2n - 3n^{2} - 3n + n

= 4n^{3} + 3n^{2}

Hence the sum of n terms 4n^{3} + 3n^{2}

(ii) 8 terms

= 4n^{3} + 3n^{2}

n = 8

= 4(8)^{3} + 3(8)^{2}

= 2048 + 192

= 2240

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