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3-1-3-5-7-2n-1-




Question Number 200718 by hardmath last updated on 22/Nov/23
3.  1+3+5+7+...+(2n+1) = ?
$$\mathrm{3}. \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+…+\left(\mathrm{2}{n}+\mathrm{1}\right)\:=\:? \\ $$
Answered by AST last updated on 22/Nov/23
2+4+..+2n=2(1+2+3+...+n)=n^2 +n  1+2+3+...+2n+1=(2n+1)(n+1)  1+3+5+...+2n+1=(n+1)(2n+1−n)=(n+1)^2
$$\mathrm{2}+\mathrm{4}+..+\mathrm{2}{n}=\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}\right)={n}^{\mathrm{2}} +{n} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{2}{n}+\mathrm{1}=\left(\mathrm{2}{n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\mathrm{2}{n}+\mathrm{1}=\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}−{n}\right)=\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by mr W last updated on 22/Nov/23
S=Σ_(k=0) ^n (2k+1)  =2Σ_(k=0) ^n k+n+1  =2×((n(n+1))/2)+n+1  =(n+1)^2
$${S}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}+{n}+\mathrm{1} \\ $$$$=\mathrm{2}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n}+\mathrm{1} \\ $$$$=\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by Mathspace last updated on 23/Nov/23
1+3+5+7+....+2n+1  2n+1+2n−1+2n−3+....  Σ=2n+2+2n+2+2n+2+...(n+1)×  =2(n+1)^2 =2s ⇒  s=(n+1)^2
$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+….+\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{2}{n}+\mathrm{1}+\mathrm{2}{n}−\mathrm{1}+\mathrm{2}{n}−\mathrm{3}+…. \\ $$$$\Sigma=\mathrm{2}{n}+\mathrm{2}+\mathrm{2}{n}+\mathrm{2}+\mathrm{2}{n}+\mathrm{2}+…\left({n}+\mathrm{1}\right)× \\ $$$$=\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{s}\:\Rightarrow \\ $$$${s}=\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

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