Powers of Natural Numbers
David W. Hansen
© 2003
In the article Cubes as Sums of Consecutive Odd Integers [1], we dealt solely with writing cubes of natural numbers as sums of consecutive odd integers. But, what about other powers of natural numbers? Can they also be written as sums of consecutive odd integers? Let’s see.
54 = 625 = 5(53) = 5 addends of 53 = 53 + 53 + 53 + 53 + 53
= (53 – 4) + (53 – 2) + 53 + (53 + 2) + (53 + 4)
= 121 + 123 + 125 + 127 + 129,
a sum of 5 consecutive odd integers.
Also, 54 = 625 = 52(52) = 25 addends of 25
= 25 + . . . + 25 + 25 + 25 + . . . + 25
= (25 –24) + . . . + (25 – 2) + 25 + (25 + 2) + . . . + (25 + 24)
= 1 + 3 + 5 + 7 + . . . + 25 + . . . + 43 + 45 + 47 + 49,
a sum of 25 consecutive odd integers.
Furthermore, 86 = 8(85) = 8 addends of 85 = 85 + 85 + 85 + 85 + 85 + 85 + 85 + 85
= (85 – 7) + (85 – 5) + (85 – 3) + (85 – 1) + (85 + 1) + (85 + 3) + (85 + 5) + (85 + 7)
= 32,761 + 32,763 + 32,765 + 32,767 + 32,769 + 32,771 + 32,773 + 32,775,
a sum of 8 consecutive odd integers.
In general, if nq is any power of n, where n and q are both natural numbers greater than 1, then
nq = n(nq – 1) = nq – 1 + . . . + nq – 1, (1)
which is a sum of n addends of nq – 1. Now, if n is odd, then nq - 1 is odd, no matter whether q is even or odd, since nq – 1 is a product of odd integers. Thus, both n and nq – 1 have the same parity. Similarly, if n is even, then nq - 1 is even, no matter whether q is even or odd, since nq – 1 is a product of even integers. Thus, both n and nq – 1 have the same parity.
Then, by Theorem 1 of [2], (1) above can be represented as a sum of consecutive odd integers since it is written as a product N of two factors, n and nq - 1, each greater than 1 and of the same parity. And also from Theorem 1, we have nq = ka = n(nq – 1), so the first term F and the last term L of this sum are
F = a - (k -1) = nq – 1 – (n -1), and L = a + (k -1) = nq – 1 + (n -1),
giving us
Theorem 1
Any power of a natural number, nq, where n and q are both natural numbers
greater than 1, may be written as a sum of n consecutive odd integers.
a) The first term of this sum is F = nq – 1 – (n - 1).
b) The last term of this sum is L = nq – 1 + (n - 1).
c) Each term of the series is 2 greater than its preceding term.
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Example Find a) the first and last terms of a sum of consecutive odd integers which represent 47. Then, b) write out each term of this sum, and c) find any other sums of consecutive odd integers representing this number.
a) Since nq = 47, we have n = 4 and q = 7. By Theorem 1, the first and last terms are
F = 47 – 1 - (4 - 1) = 46 – 3 = 4093,
and L = 47 – 1 + (4 - 1) = 46 + 3 = 4099.
b) There will be n = 4 terms in this sum, and we get 47 = 4093 + 4095 + 4097 + 4099.
c) To find any other sums, we use Theorem 1 from [2]. Thus,
47 = 42(45) = ka, F = a – (k – 1) = 45 – (42 – 1) = 1024 – 15 = 1009,
L = a + (k – 1) = 45 + (42 – 1) = 1024 + 15 = 1039.
= 1009 + 1011 + 1013 + 1015 + . . . + 1033 + 1035+ 1037 + 1039. (k = 42 = 16 terms)
= 43(44) = ka, F = a – (k – 1) = 44 – (43 – 1) = 256 – 63 = 193,
L = a + (k – 1) = 44 + (43 – 1) = 256 + 63 = 319.
= 193 + 195 + 197 + 199 + . . . + 313 + 315+ 317 + 319. (k = 43 = 64 terms)
We do not need to consider 44(43), 45(42), or 46(4), since commuting the factors of nq do not give us
any new sums. See [2].
Now, to find all the various sums of consecutive odd integers which represent nq, we must examine each of the two factors, nr and ns, whose product equals nq; that is
nq = nr ns, where r = 1, 2, . . . , q -1, s = q -1, q - 2,. . . ,1, and r + s = q.
Looking at the examples below, we see that at some point, either 1) s = r, or 2) s = r + 1,
1) 46 = 41 45 = 42 44 = 43 43 = 44 42 = 45 41
2) 47 = 41 46 = 42 45 = 43 44 = 44 43 = 45 42 = 46 41
and beyond this point, we get no new sums since, in general nr ns = ns nr. Now, in
1) where q is even, r + s = q, and s = r, we get r + r = 2r = q, or r = q/2, and in
2) where q is odd, r + s = q, and s = r + 1, we get r + (r + 1) = 2r + 1 = q, or r = (q – 1)/2.
These two values of r from 1) and 2) tell us the number of different sums of consecutive odd integers we can
find to represent nq, as follows
Theorem 2
If N = nq, where n and q are both natural numbers greater than 1, then the
number of ways to represent nq as a sum of consecutive odd integers is
1) q/2 if q is even, and 2) (q – 1)/2 if q is odd.
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Example Find the number of ways in which 311 may be written as a sum of consecutive odd integers. Then, find all these sums.
Here, nq = 311, so n = 3, and q = 11. Since q = 11 is odd, there will be (q – 1)/2 = (11- 1)/2 = 5 different ways to write 311 as a sum of consecutive odd integers. They are:
1) 311 = 3(310) = [ 310 – (3 - 1)] + . . . + [ 310 + (3 - 1)]
= 59047 + 59049 + 59051 (3 terms)
2) 311 = 32(39) = [ 39 – (32 - 1)] + . . . + [ 39 + (32 - 1)]
= (19683 - 8) + . . . + (19683 + 8)
= 19,675 + 19,677 + . . . + 19,689 + 19,691 (9 terms)
3) 311 = 33(38) = [ 38 – (33 - 1)] + . . . + [ 38 + (33 - 1)]
= (6561 - 26) + . . . + (6561 + 26)
= 6,535 + 6,537 + . . . + 6,585 + 6,587 (27 terms)
4) 311 = 34(37) = [ 37 – (34 - 1)] + . . . + [ 37 + (34 - 1)]
= (2187 - 80) + . . . + (2187 + 80)
= 2,107 + 2,109 + . . . + 2,265 + 2,267 (81 terms)
5) 311 = 35(36) = [ 36 – (35 - 1)] + . . . + [ 36 + (35 - 1)]
= (729 - 242) + . . . + (729 + 242)
= 487 + 489 + . . . + 969 + 971 (243 terms)
Since each of these five sums are arithmetic series, we know that their sum S is given by
S = n (a1 + an) / 2, where n is the number of terms, a1 is the first term, and a2 is the last term.
We may use this formula to check our answers. For example, in 5) above,
487 + 489 + . . . + 969 + 971 = 243( 487 + 971 )/2
= 243 (1458)/2 = 243(729) = 35(36) = 311.
References
(mathematics1.fortunecity.com/CubesAsSums.html)
(mathematics1.fortunecity.com/ConsecutiveOddIntegers.html)
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