Consider the numerical pyramid below, formed by simply putting down the series of odd numbers into a pyramid.
1
3 5
7 9 11
13 15 17 19
. . .
Find a formula for the sum of the numbers in the nth row, and prove it.
There are n elements in the nth row. The total number of elements in the previous n-1 rows is
n(n-1)/2; the first element in row n is the next odd number, given by 2(n(n-1)/2 +1)-1
i.e. n^2-n+1
Each odd number in the row increases by 2, giving a total of
n(n^2-n+1)+2(n(n-1)/2 = n^3-n^2-n+n^2-n = n^3
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Posted by DrBob
on 2003-10-19 12:04:53 |