¿Por qué es $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$?

Question

Podría alguien por favor ampliar sobre cómo obtener de $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ $\;\left(\dfrac{n^2}{2}\right)\;?\;$

Me parece que no puede envolver mi cabeza alrededor de eso.

Answer 1

Esto se reduce a probar que $1-\frac 12 = \frac 12$. Para ello, expanda $1$ en la serie geométrica $$1=\sum_{i=1}^{\infty}\frac 1{2^i},$$ and divide both sides by $2$ to get $$\frac 12=\sum_{i=1}^{\infty}\frac 1{2^{i+1}}=\sum_{i=2}^{\infty}\frac 1{2^i}.$$ It follows that $$1-\frac 12 = \sum_{i=1}^{\infty}\frac 1{2^i}-\frac 1{2^1}=\sum_{i=2}^{\infty}\frac 1{2^{i}}=\frac 12.$$

Answer 2

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Answer 3

$$n^2 = \dfrac{2n^2}{2}$$ $$\dfrac{2 n^2}{2} - \dfrac{n^2}{2} = \dfrac{2n^2 - n^2}{2}= \dfrac{n^2}{2}$$

En resumen, para cualquier $X$: $\;(X = x; X = n^2; \text{ or}\;X = n^{917})\;\text{etc.}:\;$ $${\bf two}\text{ halves of X}\;- \;{\bf one} \text{ half of X}\; = \;{\bf one} \text{ half of X}$$

Answer 4

Procedemos por inducción sobre $n$. El caso base para$n = 0$$0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, lo que se deduce del hecho de que $0$ es la identidad aditiva en $\mathbb{Q}$. Suponga que la demanda se mantiene para algunos $n = k$. Ahora \begin{aligned} (k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} - \frac{1}{2} \\ &= \color{blue}{\left ( k^2 - \frac{k^2}{2} \right )} + k + \frac{1}{2} \\ &= \color{blue}{\frac{k^2}{2}} + k + \frac{1}{2} \quad \color{blue}{\text{(hypothesis)}} \\ &= \frac{k^2 + 2k + 1}{2} \\ &= \frac{(k+1)^2}{2}. \end{aligned} Esto concluye el paso inductivo, por lo que nuestro reclamo tiene para todos los $n \in \mathbb{N}$.