Question Number 39519 by math khazana by abdo last updated on 07/Jul/18
$${simplify}\: \\ $$$$\left.\mathrm{1}\right)\:{A}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{{a}}}\left\{\:\left(\frac{\mathrm{1}+\sqrt{{a}}}{\mathrm{2}}\:\right)^{{n}} \:−\left(\frac{\mathrm{1}−\sqrt{{a}}}{\mathrm{2}}\right)^{{n}} \right\}\:{with}\:{n}\:{natural}\: \\ $$$${integr}\:{and}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}}\left\{\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} \:−\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$
Commented by math khazana by abdo last updated on 08/Jul/18
![1) we have A_n = (1/(2^n (√a))){ (1+(√a))^n −(1−(√a))^n } = (1/(2^n (√a))){ Σ_(k=0) ^n C_n ^k ((√a))^k −Σ_(k=0) ^n C_n ^k (−(√a))^k } = (1/(2^n (√a))){ Σ_(k=0) ^n C_n ^k ( ((√a))^k −(−(√a))^k )} =(1/(2^n (√a))){ Σ_(k=0) ^n C_n ^k (1−(−1)^k )((√a))^k } = (1/(2^n (√a))) Σ_(p=0) ^([((n−1)/2)]) 2 C_n ^(2p+1) ((√a))^(2p+1) } A_n = (1/2^(n−1) ) Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) a^p 2) let take a=2x+1 ⇒ f(x) = (1/2^(n−1) ) Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (2x+1)^p .](https://www.tinkutara.com/question/Q39589.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{\:\left(\mathrm{1}+\sqrt{{a}}\right)^{{n}} \:−\left(\mathrm{1}−\sqrt{{a}}\right)^{{n}} \right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left(\sqrt{{a}}\right)^{{k}} \:\:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−\sqrt{{a}}\right)^{{k}} \right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\left(\:\left(\sqrt{{a}}\right)^{{k}} \:−\left(−\sqrt{{a}}\right)^{{k}} \right)\right\} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(\mathrm{1}−\left(−\mathrm{1}\right)^{{k}} \right)\left(\sqrt{{a}}\right)^{{k}} \right\} \\ $$$$\left.\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:\left(\sqrt{{a}}\right)^{\mathrm{2}{p}+\mathrm{1}} \right\} \\ $$$${A}_{{n}} =\:\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:{a}^{{p}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{take}\:{a}=\mathrm{2}{x}+\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{{p}} \:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18
$$\left.\mathrm{1}\right){A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{\left(\mathrm{1}+{n}_{{C}_{\mathrm{1}} } \left(\sqrt{{a}}\right)+{n}_{{C}_{\mathrm{2}} } \left(\sqrt{{a}}\right)^{\mathrm{2}} +{n}_{{C}_{\mathrm{3}} } \left(\sqrt{{a}}\right)^{\mathrm{3}} +\right.\right. \\ $$$$\left.\:\left.+…+\left(\sqrt{{a}}\right)^{{n}} \right)−\left(\mathrm{1}−{n}_{{C}_{\mathrm{1}} } \left(\sqrt{{a}}\right)^{\mathrm{2}} +{n}_{{C}_{\mathrm{2}} } \left(\sqrt{{a}}\right)^{\mathrm{3}} −…\right)\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}^{{n}} \sqrt{{a}}}\left\{{n}_{{C}_{\mathrm{1}} } \left(\sqrt{{a}}\right)+{n}_{{C}_{\mathrm{3}} } \left(\sqrt{{a}}\right)^{\mathrm{3}} +{n}_{{C}_{\mathrm{5}} } \left(\sqrt{{a}}\right)^{\mathrm{5}} +..\right. \\ $$