Question
Answer
$$(n-i+1)^2=i^2-2in+n^2-2i+2n+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-i+1)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}i^2-2in+n^2-2i+2n+1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{n-i+1}\right) $ by each term in $ \left( n-i+1\right) $. $$ \left( \color{blue}{n-i+1}\right) \cdot \left( n-i+1\right) = n^2-in+n-in+i^2-i+n-i+1 $$ |
| ② | Combine like terms: $$ n^2 \color{blue}{-in} + \color{red}{n} \color{blue}{-in} +i^2 \color{green}{-i} + \color{red}{n} \color{green}{-i} +1 = i^2 \color{blue}{-2in} +n^2 \color{green}{-2i} + \color{red}{2n} +1 $$ |
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