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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 deleteduser1 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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