Let M = sum of the cubes of the first 2024 odd numbers
and N = sum of the cubes of the first 2023 even numbers.
Find M-N
Sums of odd and even cubes, up to n:
n^2(2n^2-1) (odd)
2n^2(n+1)^2 (even)
So, for (n-1) even numbers: 2(n-1)^2n^2
Deducting:
n^2(2n^2-1)-2(n-1)^2n^2
=n^2((2n^2-1)-2(n-1)^2), when (2n^2-1)-2(n-1)^2=(4n-3)
=n^2(4n-3)
Let n=2024 then n^2(4n-3) = 33,153,589,568
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Posted by broll
on 2023-11-03 07:20:52 |