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I'm trying to understand summation for my algorithm course and it has been a while since I took discrete math. Could any body please explain how does summation simplification work from the problem below? How did it got the result? a detailed step-by-step explanation will help me a lot. Thank you.

$$\sum_{i=1}^n (n-i+1) = \frac12n(n+1)$$

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You may observe that $$ \sum_{i=1}^n i=\sum_{i=1}^n (n+1-i), \qquad (i \to n+1-i) \tag1 $$ giving $$ \sum_{i=1}^n i=\sum_{i=1}^n (n+1)-\sum_{i=1}^n i $$ or $$ \begin{align} 2 \times\sum_{i=1}^n i&=(n+1)\sum_{i=1}^n 1=(n+1)n \end{align} $$ to obtain $$ \sum_{i=1}^n i=\frac{n(n+1)}2, $$ you conclude using $(1)$.

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