Following Gauss's Path
It is said that, as a child, Gauss completed the tedious assignment of adding the first one hundred natural numbers in a few seconds by realizing that 1 + 1 0 0 = 2 + 9 9 = … = 5 0 + 5 1 , so the answer must be 5 0 × 1 0 1 , since there are 50 such couples. The general case can be written as: k = 1 ∑ n k = 2 n ( n + 1 )
So, the sum of the first n natural numbers is a 2nd-degree polynomial.
It can be shown that 1 p + 2 p + … + n p = a 1 n + a 2 n 2 + … + a p n p + a p + 1 n p + 1 , a polynomial of degree p + 1 .
What is the value of a 1 + a 2 + … + a p + 1 ?
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1 solution
But does it work for n = 1 as well?
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That certainly is this problem's bonus. By substituting n for 1 , 2 , 3 … p and ( p + 1 ) , you can build a system of linear equations to determine the coefficients.
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No, my point here is that you have only demonstrated that the sum is 1 for a finite number of n 's. How do you know that it works for n = 1 , 2 , 3 , 4 , … ?
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@Pi Han Goh – I don't know. Can you illuminate me?
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@Gabriel Chacón – I think there is a typo in Pi Han Goh's last reply. I think he meant if it worked for n<>1, 2, 3, 4... I suppose he points to the intermediate (decimal) values of n.
I tried the shortcut by thinking on the n(n+1)/2 = (n^2+n)/2 formula but only considered the numerator, so I got the sum to be 2 and the answer for the general case to be (probably) equal to p+1. I failed since I missed the denominator. (It doesn't mean I could otherwise have reached the right answer).
Notice that when n = 1 , the sum k = 1 ∑ n k p has only one term: 1 p = 1 . So 1 should be the value the polynomial yields when we substitute n for 1.
Therefore, a 1 + a 2 + … + a p + 1 = 1 .