Consider the numerical pyramid below, formed by simply putting down the series of odd numbers into a pyramid.
7 9 11
13 15 17 19
. . .
Find a formula for the sum of the numbers in the nth row, and prove it.
(In reply to Puzzle Solution
by K Sengupta)
Let G(x) = Sum of all the numbers from the first row to the nth
g(x) = sum of all the numbers in the xth row.
S(x) = sum to x terms of the series (1+3+5+..)
Now, we observe that the xth term of the series 1+3+5... is (2x-1)
Thus, S(x) = (x/2)(1+2x-1) = x^2
Now, G(n) = S(1+2+...+n) = S[(n^2+n)/2) = n^2*(n+1)^2/4,
and, G(n-1) = S(1+2+...+(n-1)) = S[(n^2-n)/2) = n^2*(n-1)^2/4.
g(n) = G(n) - G(n-1) = (n^2)/4*[(n+1)^2 - (n-1)^2] = (n^2)/4*(4n) = n^3
Therefore, the required sum of the numbers in the nth row is n^3.