Question Number 46136 by Meritguide1234 last updated on 21/Oct/18
Answered by ajfour last updated on 21/Oct/18
![(cos nx)^(1/n) = (1−2sin^2 ((nx)/2))^(1/n) if lim_(x→0) then = 1−(2/n)(((n^2 x^2 )/4)) = 1−((nx^2 )/2) L = (1/2)[((n(n+1))/2)−1] L = ((n^2 +n−2)/4)](https://www.tinkutara.com/question/Q46138.png)
$$\left(\mathrm{cos}\:{nx}\right)^{\mathrm{1}/{n}} \:=\:\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \:\frac{{nx}}{\mathrm{2}}\right)^{\mathrm{1}/{n}} \\ $$$$\:\:\:\:\:{if}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:{then}\:\:=\:\mathrm{1}−\frac{\mathrm{2}}{{n}}\left(\frac{{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:=\:\mathrm{1}−\frac{{nx}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${L}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{1}\right]\: \\ $$$$\:\:\:{L}\:=\:\frac{{n}^{\mathrm{2}} +{n}−\mathrm{2}}{\mathrm{4}}\: \\ $$
Commented by Meritguide1234 last updated on 21/Oct/18
$${nice} \\ $$
Commented by Meritguide1234 last updated on 21/Oct/18
Commented by rahul 19 last updated on 21/Oct/18
![Ajfour sir, pls explain this last line. How you got L=(1/2)[n(n+1)/2 −1]..??](https://www.tinkutara.com/question/Q46148.png)
$${Ajfour}\:{sir},\:{pls}\:{explain}\:{this}\:{last}\:{line}. \\ $$$${How}\:{you}\:{got}\:{L}=\frac{\mathrm{1}}{\mathrm{2}}\left[{n}\left({n}+\mathrm{1}\right)/\mathrm{2}\:−\mathrm{1}\right]..?? \\ $$
Commented by Meritguide1234 last updated on 21/Oct/18
