Question Number 199838 by Calculusboy last updated on 10/Nov/23
Answered by Frix last updated on 10/Nov/23
$${y}=\sqrt[{{x}}]{{xy}} \\ $$$${y}={x}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\left({x}+\mathrm{5}\right)^{{x}} =\mathrm{23}{x}+\mathrm{1}+{x}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\mathrm{Trying}\:{x}=\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\underset{{i}=−\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{4}{i}+\mathrm{7}}{\mathrm{4}^{{i}+\mathrm{1}} }\:=\frac{\mathrm{52}}{\mathrm{9}}−\frac{\left(\mathrm{12}{n}+\mathrm{37}\right)}{\mathrm{4}^{{n}} \mathrm{36}} \\ $$$${E}=\underset{{i}=−\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{4}{i}+\mathrm{7}}{\mathrm{4}^{{i}+\mathrm{1}} }\:=\frac{\mathrm{52}}{\mathrm{9}} \\ $$
Commented by Calculusboy last updated on 11/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$