How to prove that $\displaystyle 1+3+5+...+n = \frac{(n+1)^2}{4}$ ?
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This is an arithmetic seriesHow to prove that: $\displaystyle 1+3+5+...+n = \frac{(n+1)^2}{4}$ ?
. . with first term $\displaystyle a = 1$, common difference $\displaystyle d = 2$, and $\displaystyle \frac{n+1}{2}$ terms
Using the Sum formula: . $\displaystyle S \:= \:\frac{1}{2}\cdot\frac{n+1}{2}\left[2\cdot1 + 2\cdot\left(\frac{n+1}{2} - 1\right)\right] $
. . . $\displaystyle = \;\frac{n+1}{4}[2 + (n + 1 - 2)] \;= \;\frac{n+1}{4}(n+1) \;= \;\frac{(n+1)^2}{4}$