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Question-57617

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Question Number 57617 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
source Hall andknight algebra
$${source}\:{Hall}\:{andknight}\:\:{algebra} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
28)(a_1 /(1+a_1 ))=1−(1/(1+a_1 )) (a_2 /((1+a_1 )×(1+a_2 )))−(1/(1+a_1 ))=((a_2 −1−a_2 )/((1+a_1 )(1+a_2 )))=−(1/((1+a_1 )(1+a_2 ))) (a_3 /((1+a_1 )(1+a_2 )(1+a_3 )))−(1/((1+a_1 )(1+a_2 )))=−(1/((1+a_1 )(1+a_2 )(1+a_3 ))) .... .... now add them so required sum is S=1−(1/((1+a_1 )(1+a_2 )(1+a_3 )....(1+a_n )))
$$\left.\mathrm{28}\right)\frac{{a}_{\mathrm{1}} }{\mathrm{1}+{a}_{\mathrm{1}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{1}} } \\ $$$$\frac{{a}_{\mathrm{2}} }{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)×\left(\mathrm{1}+{a}_{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{1}+{a}_{\mathrm{1}} }=\frac{{a}_{\mathrm{2}} −\mathrm{1}−{a}_{\mathrm{2}} }{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)} \\ $$$$\frac{{a}_{\mathrm{3}} }{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\left(\mathrm{1}+{a}_{\mathrm{3}} \right)}−\frac{\mathrm{1}}{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\left(\mathrm{1}+{a}_{\mathrm{3}} \right)} \\ $$$$…. \\ $$$$…. \\ $$$${now}\:{add}\:{them} \\ $$$${so}\:{required}\:{sum}\:{is} \\ $$$${S}=\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\left(\mathrm{1}+{a}_{\mathrm{3}} \right)….\left(\mathrm{1}+{a}_{{n}} \right)} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
26)(1/(1+x))+(1/(1−x))=(2/(1−x^2 )) (2/(1+x^2 ))+(2/(1−x^2 ))=(4/(1−x^4 )) (4/(1+x^4 ))+(4/(1−x^4 ))=(8/(1−x^8 )) ..... .... add them S+(1/(1−x))=(2^n /(1−x^2^n )) S=−(1/(1−x))+(2^n /(1−x^2^n ))
$$\left.\mathrm{26}\right)\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}=\frac{\mathrm{2}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{4}}{\mathrm{1}−{x}^{\mathrm{4}} }=\frac{\mathrm{8}}{\mathrm{1}−{x}^{\mathrm{8}} } \\ $$$$….. \\ $$$$…. \\ $$$${add}\:{them}\: \\ $$$$\boldsymbol{{S}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}=\frac{\mathrm{2}^{{n}} }{\mathrm{1}−{x}^{\mathrm{2}^{{n}} } } \\ $$$${S}=−\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{2}^{{n}} }{\mathrm{1}−{x}^{\mathrm{2}^{{n}} } } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
27) (x/(1−x^2 ))+(1/(1−x^2 ))=((x+1)/(1−x^2 ))=(1/(1−x)) (x^2 /(1−x^4 ))+(1/(1−x^4 ))=((x^2 +1)/(1−x^4 ))=(1/(1−x^2 )) that means (x/(1−x^2 ))=(1/(1−x))−(1/(1−x^2 )) (x^2 /(1−x^4 ))=(1/(1−x^2 ))−(1/(1−x^4 )) ... ... (x^2^(n−1) /(1−x^2^n ))=(1/(1−x^2^(n−1) ))−(1/(1−x^2^n )) add them S=(1/(1−x))−(1/(1−x^2^n ))
$$\left.\mathrm{27}\right) \\ $$$$ \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{4}} }=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${that}\:{means} \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$$$… \\ $$$$… \\ $$$$\frac{{x}^{\mathrm{2}^{{n}−\mathrm{1}} } }{\mathrm{1}−{x}^{\mathrm{2}^{{n}} } }=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}^{{n}−\mathrm{1}} } }−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}^{{n}} } } \\ $$$${add}\:{them} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}^{{n}} } } \\ $$$$ \\ $$$$ \\ $$

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