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