How would u prove this? (1 Viewer)

eternallyboreduser · Dec 23, 2023

kendricklamarlover101 said:
i just realised this is invalid since u cant apply the binomial theorem once again...
well another approach u could do is RHS-LHS
$RHS-LHS = \frac{3^k}{2^{2k}} - \frac{1}{2^{k+1}} - \frac{1}{2}$
$=\frac{3^k - 2^{k-1} - 2^{2k-1}}{2^{2k}}$
$=\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}}$
Note that $3^k$ and $2^{k}$ are increasing for all real numbers k
Since 0<k<1, $3^k > 3^0 = 1$ and similarily, $1+2^k>1+2^0=2$
Also, -1<k-1<0, so $2^{k-1} > 2^{-1} = \frac{1}{2}$
Hence $\frac{3^k -2^{k-1}\left(1+2^k\right)}{2^{2k}} > \frac{1 -\frac{1}{2}\left(2\right)}{2^{2k}}=\frac{0}{2^{2k}}$
$\therefore RHS-LHS>0 \implies LHS < RHS$
surely i didnt make another error right

Just saw this lol but thanks

How would u prove this? (1 Viewer)

eternallyboreduser

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