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Question Number 194732 by MathsFan last updated on 14/Jul/23

Commented by York12 last updated on 15/Jul/23

do not participate IYMC it is fake

$${do}\:{not}\:{participate}\:{IYMC} \\ $$$${it}\:{is}\:{fake} \\ $$

Answered by mahdipoor last updated on 14/Jul/23

A⇒f(x)=(π^x −(1/π)).x^n .((1/π^2 )−π^x ) ⇒f=0⇒x=−1,0,−2 B⇒2^n −(−1)^n ≡ { ((2^(2m+1) −(−1)^(2m+1) )),((2^(2m) −(−1)^(2m) )) :}≡ { ((2.4^m +1)),((4^m −1)) :}≡ { ((2+1)),((1−1)) :}≡0 , mod(3) m∈W C⇒inside Σ=((1+(√2))/(1+2.((√2)/2)))−(8/((2)^2 ))=−1 ,Σ_0 ^(10) (−1)=1 log_3 (((sin2π)/(...))+1)=log_3 1=0

$${A}\Rightarrow{f}\left({x}\right)=\left(\pi^{{x}} −\frac{\mathrm{1}}{\pi}\right).{x}^{{n}} .\left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }−\pi^{{x}} \right) \\ $$$$\Rightarrow{f}=\mathrm{0}\Rightarrow{x}=−\mathrm{1},\mathrm{0},−\mathrm{2} \\ $$$${B}\Rightarrow\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} \equiv\begin{cases}{\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} −\left(−\mathrm{1}\right)^{\mathrm{2}{m}+\mathrm{1}} }\\{\mathrm{2}^{\mathrm{2}{m}} −\left(−\mathrm{1}\right)^{\mathrm{2}{m}} }\end{cases}\equiv\begin{cases}{\mathrm{2}.\mathrm{4}^{{m}} +\mathrm{1}}\\{\mathrm{4}^{{m}} −\mathrm{1}}\end{cases}\equiv \\ $$$$\begin{cases}{\mathrm{2}+\mathrm{1}}\\{\mathrm{1}−\mathrm{1}}\end{cases}\equiv\mathrm{0}\:\:\:,\:\:{mod}\left(\mathrm{3}\right)\:\:\:\:{m}\in{W}\:\:\:\: \\ $$$${C}\Rightarrow{inside}\:\Sigma=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{1}+\mathrm{2}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}−\frac{\mathrm{8}}{\left(\mathrm{2}\right)^{\mathrm{2}} }=−\mathrm{1}\:,\underset{\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)=\mathrm{1} \\ $$$${log}_{\mathrm{3}} \left(\frac{{sin}\mathrm{2}\pi}{...}+\mathrm{1}\right)={log}_{\mathrm{3}} \mathrm{1}=\mathrm{0} \\ $$

Commented by York12 last updated on 15/Jul/23

bro it is an onlie exam called IYMC & he is cheating

$${bro}\:{it}\:{is}\:{an}\:{onlie}\:{exam}\: \\ $$$${called}\:{IYMC}\:\&\:{he}\:{is}\:{cheating} \\ $$

Answered by qaz last updated on 15/Jul/23

2^n −(−1)^n ≡(3−1)^n −(−1)^n ≡(−1)^n −(−1)^n =0(mod 3) ⇒3∣2^n −(−1)^n

$$\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} \equiv\left(\mathrm{3}−\mathrm{1}\right)^{{n}} −\left(−\mathrm{1}\right)^{{n}} \equiv\left(−\mathrm{1}\right)^{{n}} −\left(−\mathrm{1}\right)^{{n}} =\mathrm{0}\left({mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{3}\mid\mathrm{2}^{{n}} −\left(−\mathrm{1}\right)^{{n}} \\ $$

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