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.
obviously the sum of the numbers in row n is given by n^3.
Proof:
the first number in the nth row is given by
n*(n-1)+1
This is cos till the end of the n-1th row, the number of odd numbers that would have been used up is given by 1+2+3+4....n-1 i.e n(n-1)/2 .. if that many odd numbers are used up, the last number in the n-1th row is
(n(n-1)/2 *2 -1) ..so the first number of the nth row is n(n-1)/2 *2 -1 +2 = n(n-1)+1
There are n odd numbers in row n starting with
n(n-1)+1, n(n-1)+1+2, n(n-1)1+4 ..n(n-1)+1+2(n-1)
The sum can be restated as
n*(n(n-1)+1))+2*(1+2+3+4.....n-1)
= n(n(n-1)+1)+2*(n(n-1)/2
= n(n(n-1)+1)+n(n-1)
= n^2(n-1)+n+n(n-1)
= n^3-n^2+n+n^2-n = n^3 q.e.d
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Posted by anand
on 2003-10-16 04:54:24 |