I'm trying to find the value of the following:
$(1^2 +3^2 + 5^2 + \ldots + 99^2)-(2^2 + 4^2 + 6^2 + \ldots + 100^2)+(4+8+12 + \ldots +200)$
Got the last part of the question, but I'm stuck on the first two parts.
Can anyone help me out with this?
2026-03-28 15:11:55.1774710715
Determine the value $(1^2 +3^2 + \ldots 99^2)-(2^2 + 4^2 +\ldots 100^2)+(4+8+\ldots200)$?
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Reorder the squares as $(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...(99^2-100^2)=-(3+7+11+...+199)$. Then the required quantity will be $(4-3)+(8-7)+...+(200-199)=1+1+1+...+1$ and you are to count the number of $1$'s.