# Summation notation of X^2 [duplicate]

$$\sum_{k=1}^x(k + k - 1) = x^2$$

WolframAlpha

• Remember what Sum_{k=1}^x k is. Use it and calculate, Mathematica not needed here. – mgamer Oct 28 '17 at 21:17
• Awesome! Thanks mgamer for bringing it down to basics. x^2/2 + x/2 + x^2/2 + x/2 -x = x^2 – ID10T_ERROR Oct 28 '17 at 21:59

The base case is that $\sum_{k=1}^1(2k - 1) = 1^2$ and this is pretty much self evident. Supposing it holds for the first $n$ positive integers means that:
$$\sum_{k=1}^{n+1}(2k - 1) = \sum_{k=1}^{n}(2k - 1) +2(n+1)-1=n^2+2n+1=(n+1)^2$$