1^3+2^3+3^3+...................+n^3=?
Answers
Answer:
To answer a number of similar questions, I can give the general formula, which allows one to make all sums he can wish of the powers k of the natural numbers from 1 to n. The basic idea is fairly simple: most of the effort, as we shall see, goes into giving an “elegant” appearance to the general formula.
Frankly, I am surprised to find on the Web that one generally prefers to refer to the solution of the general problems as given by the so-called “Faulhaber formulas” (1631), given in the form:
Although frightening enough, the formula just represents polynomials of degree p+1, where B(j) are the Bernoulli number, which were discovered in the early XVIII century. Faulhaber, in fact, gave the polynomials up to p =17, not the general formula.
Our approach will be much simpler:
We start from the formula of the power k+1 of a binomial (1+x), using the binomial coefficients, for example as they result from the Tartaglia/Pascal triangle:
We substitute for x the values 1, 2 , 3, 4, 5….,n, thus finding the following equations
Now we sum the equations term by term, having introduced the notation:
We observe that all LHS members disappear from the sum, excepting the last one, while the last terms of the RHS also disappear, excepting the first, which is 1, independent of k.
The result is:
This is our general solution, a recursive formula which provides all sums of all powers from 1 to any number n, starting with
with k=1. One should know by heart ist value (it is the famous formula found by Gauss at the age of 5), but, of course, also our general solution provides it. Let’s apply it:
and solve for the sum we are looking for.
Now we must make the effort of putting the result in the elegant, known form:
For example:
Similarly, we find that
noting that
We are now close to our objective, although we could be satisfied with the result that
in which formula the only unknown is the sum of the cubes we are looking for.
But, if we want to be elegant,
which answers to the question.
The result has been known for about two thousand years, and is also known as “Nicomachus’s Theorem”, from Nicomachus of Gerasa (60-120 CE).
I have found in the net an alternative, admirable intuitive demonstration.
Wikipedia () dixit: Many early mathematicians have studied and provided proofs of Nicomachus's theorem. Stroeker (1995) claims that "every student of number theory surely must have marveled at this miraculous fact". Pengelley (2002) finds references to the identity not only in the works of Nicomachus, but also in those of Aryabhata in India in the fifth century, and in those of Al-Karaji circa 1000 in Persia. Bressoud (2004) mentions several additional early mathematical works on this formula, by Alchabitius (Al Qabisi, tenth century, lived in present day Iraq and Syria, died in 967), Gersonides (circa 1300 France), and Nilakantha Somayaji (circa 1444-1544 Kerala, India); Bressoud reproduces Nilakantha's visual proof.
Bressoud, in his work, seems not to be aware of, or not to accept the fact that the theorem is attributed to Nicomachus, and attributes it to Aryabatha, 400 years later. The visual proof by Nilakantha he reproduces is basically the same I give below. It might be original of Nilakantha, it might not. All I can say is that, as the “demonstration” given by Nicomachus is not in graphic terms, one may attribute the intuitive demonstration here below to some of the later mathematicians, be he Greek or Arab, or Indian, or Chinese ( a candidate could be Jia Xian, of the XI century).
Here is the demonstration (from ):
The area of the square can be looked at in different ways.”
The columns are in succession 1, 2, 3, 4, 5, 6 elementary squares wide. The area of the square is thus (1+2+3+4+5+6)^2.
But we can also consider the square starting from the top left hand corner, and we see that there is one square composed of an elementary square (red); a total of two squares dark yellow, totalling 8 (=2^3) elementary squares; 3 yellow squares, totalling 27 (=3^3) elementary squares; a total of 4 green squares, totalling 64 (=4^3)elementary squares, and so on, up to the 6 squares totalling 6^6 elementary squares. (When the squares have a side consisting of an even number of elementary squares, note that one of them is divided in two equal rectangles).
In other words, at least up to n=6, we can intuitively see that the two ways of calculating the areas of the square must give the same result, that is
(1+2+3+4+5+6)^2 = 1+2^3+3^3+4^3+5^3+6^3.
From which the answer to the question follows: the sum of the cubes of the consecutive natural numbers is equal to the square of the (well known) sum of the consecutive natural numbers, at least from 1 to 6 , and we have no reason to think that the result will betray us for n >6.