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 - 14 --->
= 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]2, to 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]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]2 = 14400
[n(n + 1)/2]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]2 = 2025
[n(n + 1)/2]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
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 26, 24 07:41 AM
Dec 23, 24 03:47 AM
Dec 23, 24 03:40 AM