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

Commented by York12 last updated on 15/Jul/23

$d o n o t p a r t i c i p a t e I Y M C$ $i t i s f a k e$
	Answered by mahdipoor last updated on 14/Jul/23
	$A⇒f(x)=(π^x −(1/π)).x^n .((1/π^2 )−π^x ) ⇒f=0⇒x=−1,0,−2 B⇒2^n −(−1)^n ≡ { ((2^(2m+1) −(−1)^(2m+1) )),((2^(2m) −(−1)^(2m) )) :}≡ { ((2.4^m +1)),((4^m −1)) :}≡ { ((2+1)),((1−1)) :}≡0 , mod(3) m∈W C⇒inside Σ=((1+(√2))/(1+2.((√2)/2)))−(8/((2)^2 ))=−1 ,Σ_0 ^(10) (−1)=1 log_3 (((sin2π)/(...))+1)=log_3 1=0$
	$A \Rightarrow f (x) = (π^{x} - \frac{1}{π}) . x^{n} . (\frac{1}{π^{2}} - π^{x})$ $\Rightarrow f = 0 \Rightarrow x = - 1, 0, - 2$ $B \Rightarrow 2^{n} - {(- 1)}^{n} \equiv {\begin{cases} 2^{2 m + 1} - {(- 1)}^{2 m + 1} \\ 2^{2 m} - {(- 1)}^{2 m} \end{cases} \equiv {\begin{cases} 2 . 4^{m} + 1 \\ 4^{m} - 1 \end{cases} \equiv$ ${\begin{cases} 2 + 1 \\ 1 - 1 \end{cases} \equiv 0, m o d (3) m \in W$ $C \Rightarrow i n s i d e Σ = \frac{1 + \sqrt{2}}{1 + 2 . \frac{\sqrt{2}}{2}} - \frac{8}{{(2)}^{2}} = - 1, \underset{0}{\sum^{10}} (- 1) = 1$ ${l o g}_{3} (\frac{s i n 2 π}{. . .} + 1) = {l o g}_{3} 1 = 0$
	Commented by York12 last updated on 15/Jul/23

	$b r o i t i s a n o n l i e e x a m$ $c a l l e d I Y M C & h e i s c h e a t i n g$
	Answered by qaz last updated on 15/Jul/23

	$2^{n} - {(- 1)}^{n} \equiv {(3 - 1)}^{n} - {(- 1)}^{n} \equiv {(- 1)}^{n} - {(- 1)}^{n} = 0 (m o d 3)$ $\Rightarrow 3 ∣ 2^{n} - {(- 1)}^{n}$
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