Hello! Can anyone show that (2n+2-n)/2=(1+4n)/2n+1? Thanks.
O Olsi New member Joined Oct 1, 2011 Messages 3 Oct 1, 2011 #1 Hello! Can anyone show that (2n+2-n)/2=(1+4n)/2n+1? Thanks. Last edited: Oct 2, 2011
D Deleted member 4993 Guest Oct 1, 2011 #2 Olsi said: Hello! Can anyone show that 2n+2-n/2=1+4n/2n+1? Thanks. Click to expand... Repost your problem - using parenthesis (grouping symbol), so that mathematical operations can be performed in correct order.
Olsi said: Hello! Can anyone show that 2n+2-n/2=1+4n/2n+1? Thanks. Click to expand... Repost your problem - using parenthesis (grouping symbol), so that mathematical operations can be performed in correct order.
P procyon Junior Member Joined Aug 13, 2011 Messages 53 Oct 1, 2011 #3 also note that \(\displaystyle 4^n=2^{2n}\)
O Olsi New member Joined Oct 1, 2011 Messages 3 Oct 2, 2011 #4 Subhotosh Khan said: Repost your problem - using parenthesis (grouping symbol), so that mathematical operations can be performed in correct order. Click to expand... I edited it.
Subhotosh Khan said: Repost your problem - using parenthesis (grouping symbol), so that mathematical operations can be performed in correct order. Click to expand... I edited it.
burakaltr Junior Member Joined Aug 26, 2011 Messages 126 Oct 2, 2011 #5 Olsi said: Hello! Can anyone show that (2n+2-n)/2=(1+4n)/2n+1? Thanks. Click to expand... Call x=2**n therefore x+ (x**(-1)) / 2 = (1+x**2)/(2x) x+1/x = (1+x**2)/x if x!=0 then ?
Olsi said: Hello! Can anyone show that (2n+2-n)/2=(1+4n)/2n+1? Thanks. Click to expand... Call x=2**n therefore x+ (x**(-1)) / 2 = (1+x**2)/(2x) x+1/x = (1+x**2)/x if x!=0 then ?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 2, 2011 #6 Hello, Olsi! \(\displaystyle \text{Show that: }\:\dfrac{2^n + 2^{-n}}{2} \:=\:\dfrac{1+4^n}{2^{n+1}}\) Click to expand... \(\displaystyle \displaystyle \text{Multiply by }\frac{2^n}{2^n}\!:\quad \frac{2^n}{2^n}\cdot \frac{2^n + 2^{-n}}{2} \;=\;\frac{(2^n)(2^n) + (2^n)(2^{-n})}{(2^n)(2)} \;=\;\frac{4^n + 1}{2^{n+1}} \)
Hello, Olsi! \(\displaystyle \text{Show that: }\:\dfrac{2^n + 2^{-n}}{2} \:=\:\dfrac{1+4^n}{2^{n+1}}\) Click to expand... \(\displaystyle \displaystyle \text{Multiply by }\frac{2^n}{2^n}\!:\quad \frac{2^n}{2^n}\cdot \frac{2^n + 2^{-n}}{2} \;=\;\frac{(2^n)(2^n) + (2^n)(2^{-n})}{(2^n)(2)} \;=\;\frac{4^n + 1}{2^{n+1}} \)
O Olsi New member Joined Oct 1, 2011 Messages 3 Oct 2, 2011 #7 soroban said: Hello, Olsi! \(\displaystyle \displaystyle \text{Multiply by }\frac{2^n}{2^n}\!:\quad \frac{2^n}{2^n}\cdot \frac{2^n + 2^{-n}}{2} \;=\;\frac{(2^n)(2^n) + (2^n)(2^{-n})}{(2^n)(2)} \;=\;\frac{4^n + 1}{2^{n+1}} \) Click to expand... Thank you!
soroban said: Hello, Olsi! \(\displaystyle \displaystyle \text{Multiply by }\frac{2^n}{2^n}\!:\quad \frac{2^n}{2^n}\cdot \frac{2^n + 2^{-n}}{2} \;=\;\frac{(2^n)(2^n) + (2^n)(2^{-n})}{(2^n)(2)} \;=\;\frac{4^n + 1}{2^{n+1}} \) Click to expand... Thank you!
