Sum of Cubes of First n Natural Numbers

Formula to find the sum of cubes first n natural numbers :

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

Proof : 

To find (12 + 22 + 32 + ........ + n2), let us consider the identity

(x + 1)4 - x4 = 4x3 + 6x2 + 4x + 1

When x = 1,  

24 - 14 = 4(1)3 + 6(1)2 + 4(1) + 1

When x = 2,  

34 - 24 = 4(2)3 + 6(2)2 + 4(2) + 1

When x = 3,  

44 - 34 = 4(3)3 + 6(3)2 + 4(3) + 1

Continuing this, when x = n - 1,

n4 - (n - 1)4 = 4(n - 1)3 + 6(n - 1)2 + 4(n - 1) + 1

When x = n,

(n + 1)4 - n4 = 4n3 + 6n2 + 4n + 1

Adding all these equations and cancelling the terms on the Left Hand side, we get,

(n + 1)4 - 1---> 

= 4(13 + ... + n3) + 6(12 + ... + n2) + 4(1 + ... + n) + n

n4 + 4n3 + 6n2 + 4n + 1 - 1 --->

= 4(13 + ... + n3) + 6[n(n + 1)(2n + 1)]/6 + 4[n(n + 1)]/2 + n

n4 + 4n3 + 6n2 + 4n --->

= 4(13 + ... + n3) + n(n + 1)(2n + 1) + 2n(n + 1) + n

n4 + 4n3 + 6n2 + 4n --->

= 4(13 + ... + n3) + 2n3 + 3n2 + n + 2n2 + 2n + n

n4 + 4n3 + 6n2 + 4n = 4(13 + ... + n3) + 2n3 + 5n2 + 4n

4(13 + 23 + ...... + n3) = n4 + 4n3 + 6n2 + 4n - 2n3 - 5n2 - 4n

4(13 + 23 + 33 + ...... + n3) = n4 + 2n3 + n2

4(13 + 23 + 33 + ...... + n3) = n2(n2 + 2n + 1)

4(13 + 23 + 33 + ...... + n3) = n2(n + 1)2

Divide each side by 4. 

13 + 23 + 33 + ...... + n3 = [n2(n + 1)2]/4

13 + 23 + 33 + ...... + n3 = [n2(n + 1)2]/22

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

Example 1 :

Find the value of

13 + 23 + 33 + ........ + 173

Solution :

= 13 + 23 + 33 + ........ + 173

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

= [17(17 + 1)/2]2

[17(18)/2]2

= [17(9)]2

= 1532

= 23409

Example 2 : 

Find the value of

83 + 103 + 113 + ........ + 213

Solution :

83 + 103 + 113 + ........ + 213 :

= (13 + 23 + 33 + ........ + 213) - (13 + 23 + 33 + ........ + 73)

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

[21(21 + 1)/2]2 - [7(7 + 1)/2]2

[21(22)/2]2 - [7(8)/2]2

[21(11)]2 - [7(4)]2

= 2312 - 282

= 53361 - 784

= 52577

Example 3 : 

Find the value of

33 + 63 + 93 + ........ + 453

Solution :

33 + 63 + 93 + ........ + 453

= (3x1)3 + (3x2)3 + (3x3)3 + ........ + (3x15)3

= 3313 + 3323 + 3333 + ........ + 33153

= 33(13 + 23 + 33 + ........ + 153)

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

= 33[15(15 + 1)/2]2

= 33[15(16)/2]2

= 33[15(8)]2

= 27(120)2

= 27(14400)

= 388800

Example 4 : 

Find the sum of 

2 + 16 + 54 + ........ + to 12 terms

Solution :

2 + 16 + 54 + ........ + to 12 terms

=  2(1 + 8 + 27 + ........ to 12 terms)

=  2(13 + 23 + 33 + ........ to 12 terms)

=  2(12 + 22 + 32 + ........ + 122)

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

= 2[12(12 + 1)/2]2

= 2[12(13)/2]2

= 2[6(13)]2

= 2(78)2

= 2(6084)

= 12164

Example 5 : 

Find the sum of

3 + 24 + 81 + ........ + 6591

Solution :

3 + 24 + 81 + ........ + 6591 :

= 3(1 + 8 + 27 + ........ + 2197)

=  3(13 + 23 + 33 + ........ + 133)

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

= 3[13(13 + 1)/2]2

= 3[13(14)/2]2

= 3[6(7)]2

= 3(42)2

= 3(1764)

= 5292

Example 6 :

Find the average of cubes first 10 natural numbers. 

Solution :

Using 13 + 23 + 33 + ........ + n3 = [n(n + 1)/2]2to find the sum of squares first 10 natural numbers. 

13 + 23 + 33 + ........ + 103 :

= [10(10 + 1)/2]2

= [10(11)/2]2

= [5(11)]2

= 552

= 3025

Average of cubes first 10 natural numbers : 

= (Sum of cubes first 10 natural numbers)/10

= 3025/10

= 302.5

Example 7 : 

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

13 + 23 + 33 + ........ + k3

Solution :

1 + 2 + 3 + ........ + k = 325

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

k(k + 1)/2 = 325

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

13 + 23 + 33 + ........ + k3 = [k(k + 1)/2]2

Substitute k(k + 1)/2 = 325, 

13 + 23 + 33 + ........ + k3 = 3252

13 + 23 + 33 + ........ + k3 = 105625

Example 8 : 

If 13 + 23 + 33 + ........ + k3 = 44100, then find 

1 + 2 + 3 + ........ + k

Solution :

13 + 23 + 33 + ........ + k3 = 44100

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

[k(k + 1)/2]= 44100

[k(k + 1)/2]2 = 2102

k(k + 1)/2 = 210

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

1 + 2 + 3 + ........ + k = 210

Example 9 : 

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

Solution :

Let n be the number of terms to be taken to get the sum 14400.

13 + 23 + 33 + ........ to n terms = 14400

13 + 23 + 33 + ........ + n3 = 14400

[n(n + 1)/2]= 14400

[n(n + 1)/2]= 1202

n(n + 1)/2 = 120

Multiply each side by 2.

n(n + 1) = 240

n2 + n = 240

n2 + n - 240 = 0

Factor and solve. 

n2 + 16n - 15n - 240 = 0

n(n + 16) - 15(n + 16) = 0

(n + 16)(n - 15) = 0 

n + 16 = 0

n = -16

n -15 = 0

n = 15

But n ≠ -16, because number of terms can not be a negative value.

Hence n = 15.

In the given series, 15 terms to be taken to get the sum 14400.

Example 10 :

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 :

Sum of the squares of first n natural numbers is 285 :

12 + 22 + 32 + ........ + n2 = 285

[n(n + 1)(2n + 1)]/6 = 285

Multiply each side by 6.

n(n + 1)(2n + 1) = 1710 ----(1)

Sum of their cubes is 2025 :

13 + 23 + 33 + ........  + n3 = 2025

[n(n + 1)/2]= 2025

[n(n + 1)/2]= 452

n(n + 1)/2 = 45

Multiply each side by 2.

n(n + 1) = 90

Substitute n(n + 10 = 90 in (1). 

90(2n + 1) = 1710

Divide each side by 90.

2n + 1 = 19

Subtract 1 from each side. 

2n = 18

Divide each side by 2. 

n = 9

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