Question Number 136433 by metamorfose last updated on 21/Mar/21
$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{n}+\mathrm{1}} ^{\mathrm{2}{k}+\mathrm{1}} \mathrm{8}^{{k}} =…??? \\ $$
Answered by Olaf last updated on 22/Mar/21
![(2(√2)+1)^(2n+1) = Σ_(p=0) ^(2n+1) C_p ^(2n+1) (2(√2))^p (−2(√2)+1)^(2n+1) = Σ_(p=0) ^(2n+1) (−1)^p C_p ^(2n+1) (2(√2))^p (1+2(√2))^(2n+1) −(1−2(√2))^(2n+1) = = Σ_(p=0) ^(2n+1) [1−(−1)^p ]C_p ^(2n+1) (2(√2))^p = 2Σ_(k=0) ^n C_(2k+1) ^(2n+1) (2(√2))^(2k+1) =4(√2) Σ_(k=0) ^n C_(2k+1) ^(2n+1) 8^k ⇒ Σ_(k=0) ^n C_(2k+1) ^(2n+1) 8^k = (1/( 4(√2)))[(1+2(√2))^(2n+1) −(1−2(√2))^(2n+1) ]](https://www.tinkutara.com/question/Q136439.png)
$$ \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}\mathrm{C}_{{p}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{{p}} \\ $$$$\left(−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \:=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{p}} \mathrm{C}_{{p}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{{p}} \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} −\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:= \\ $$$$=\:\underset{{p}=\mathrm{0}} {\overset{\mathrm{2}{n}+\mathrm{1}} {\sum}}\left[\mathrm{1}−\left(−\mathrm{1}\right)^{{p}} \right]\mathrm{C}_{{p}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{{p}} \\ $$$$=\:\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \mathrm{8}^{{k}} \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{\mathrm{2}{k}+\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{1}} \mathrm{8}^{{k}} \:=\:\frac{\mathrm{1}}{\:\mathrm{4}\sqrt{\mathrm{2}}}\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} −\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \right] \\ $$
Commented by metamorfose last updated on 22/Mar/21
$${thnx}\:{sir} \\ $$