Sum of Squares of First n Natural Numbers

Example 1 : 

Find the value of

1² + 2² + 3² + ........ + 50²

Solution :

= 1² + 2² + 3² + ........ + 50² 

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= [50(50 + 1)(2x50 + 1)]/6

= [50(50 + 1)(100 + 1)]/6

= [50(51)(101)]/6

= [50(51)(101)]/6

= 42925

Example 2 : 

Find the value of

15² + 16² + 17² + ........ + 28²

Solution :

15² + 16² + 17² + ........ + 28² :

= (1² + 2² + 3² + ........ + 28²) - (1² + 2² + 3² + ........ + 14²)

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= [28(28 + 1)(2x28 + 1)]/6 - [14(14 + 1)(2x14 + 1)]/6

= [28(29)(56 + 1)]/6 - [14(15)(28 + 1)]/6

= [28(29)(57)]/6 - [14(15)(29)]/6

= 7714 - 1015

= 6699

Example 3 : 

Find the value of

5² + 10² + 15² + ........ + 105²

Solution :

= 5² + 10² + 15² + ........ + 105²

= (5x1)² + (5x2)² + (5x3)² + ........ + (5x21)²

= 5²1² + 5²2² + 5²3² + ........ + 5²21²

= 5²(1² + 2² + 3² + ........ + 21²)

= 25(1² + 2² + 3² + ........ + 21²)

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= 25[21(21 + 1)(2x21 + 1)]/6

= 25[21(22)(42 + 1)]/6

= 25[21(22)(43)]/6

= 82775

Example 4 : 

Find the sum of 

2 + 8 + 18 + ........ + to 30 terms

Solution :

2 + 8 + 18 + ........ + to 30 terms : 

=  2(1 + 4 + 9 + ........ to 30 terms)

=  2(1² + 2² + 3² + ........ to 30 terms)

=  2(1² + 2² + 3² + ........ + 30²)

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= 2[30(30 + 1)(2x30 + 1)]/6

= 2[30(31)(60 + 1)]/6

= 2[30(31)(61)]/6

= 18910

Example 5 : 

Find the sum of

3 + 12 + 27 + ........ + 1875

Solution :

3 + 12 + 27 + ........ + 1875 :

= 3(1 + 4 + 9 + ........ + 625)

=  3(1² + 2² + 3² + ........ + 25²)

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= 3[25(25 + 1)(2x25 + 1)]/6

= 3[25(26)(50 + 1)]/6

= 3[25(26)(51)]/6

= 16575

Example 6 :

Find the average of squares first 25 natural numbers. 

Solution :

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6, to find the sum of squares first 25 natural numbers. 

1² + 2² + 3² + ........ + 25² :

= [25(25 + 1)(2x25 + 1)]/6

= [25(26)(50 + 1)]/6

= [25(26)(51)]/6

= 5525

Average of squares first 25 natural numbers : 

= (Sum of squares first 25 natural numbers)/25

= 5525/25

= 221

Example 7 : 

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

Solution :

Total area covered by the given 15 square color papers  :

= 10² + 11² + 12² + ........ + 24²

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

Using 1² + 2² + 3² + ........ + n² = [n(n + 1)(2n + 1)]/6,

= [24(24 + 1)(2x24 + 1)]/6 - [9(9 + 1)(2x9 + 1)]/6

= [24(25)(48 + 1)]/6 - [9(10)(18 + 1)]/6

= [24(25)(49)]/6 - [9(10)(19)]/6

= 4900 - 285

= 4615

The area of 4615 cm² can be decorated with the 15 square color papers that Rekha has. 

Example 8 : 

The sum of first n natural numbers is 120, while the sum of their squares is 1240. Find the value of n.

Solution :

Sum of first n natural numbers is 120 :

1 + 2 + 3 +, ...... + n = 120

[n(n + 1)]/2 = 120

Multiply each side by 2. 

n(n + 1) = 240

Sum of their squares is 1240 : 

1² + 2² + 3² + ........ + n² = 1240

[n(n + 1)(2n + 1)]/6 = 1240

Substitute n(n + 1) = 240.

[240(2n + 1)]/6 = 1240

40(2n + 1) = 1240

Divide each side by 40.

2n + 1 = 31

Subtract 1 from each side. 

2n = 30

Divide each side by 2.

n = 15

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