How is $\frac{3+5+7 \cdots+(2n+1)}{n}=\frac{2(1+2+3+\cdots+n)+n}{n}$ true? when I multiple out, it doesn't make sense. How is $\frac{3+5+7 \cdots+(2n+1)}{n}=\frac{2+4+6+\cdots+2n)+n}{n}$ true?
How is $\frac{3+5+7 \cdots+(2n+1)}{n}=\frac{2(1+2+3+\cdots+n)+n}{n}$ true?
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sequences-and-series
discrete-mathematics
2 Answers
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The sum $2+4+6+\cdots+2n$ contains $n$ terms. So adding $n$ is the same as adding $1$ to every term. That is, $$\eqalign{(2+4+6+\cdots+2n)+\color{red}{n} &=(2+\color{red}{1})+(4+\color{red}{1})+(6+\color{red}{1})+\cdots+(2n+\color{red}{1})\cr &=3+5+7+\cdots+(2n+1)\ .\cr}$$ Now divide both sides by $n$.
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0where does the underline n in the edited comment go? – 2017-02-15
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0Why did you want an underline? I don't understand. – 2017-02-15
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0@ David I got it now, thanks. – 2017-02-15
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Because $3+5+...+(2n+1)=(1+3+...+2n+1)-1=(n+1)^2-1=n(n+2)$ and
$2(1+2+...+n)+n=2\cdot\frac{n(n+1)}{2}+n=n(n+2)$