Mathematical induction: Prove that 2^n > 3n for all n>

[owenthakkar]

hey guys!! can anyone help me with this hard question?
its to do with mathematical induction (inequalitites)
i think im missing a step in my proof and i just cant figure it out.... :?

heres the question:

Prove that 2^n>3n for all positive integers where n is greater than or equal to 4

thanks guys
any help would be great

 

[pka]

Clearly it is true for n=4, \(\displaystyle 16 = 2^4 > 12 = \left( 3 \right)\left( 4 \right)\).
So we are on the path. Suppose we can get to the Kth step, that is
\(\displaystyle 2^K > \left( 3 \right)\left( K \right)\) is true.

Now look at \(\displaystyle 2^{K+1}\).
We have \(\displaystyle \begin{array}{rcl}
2^{K + 1} & = & 2\left( {2^K } \right) \\
& > & 2\left( 3 \right)\left( K \right) \\
& = & \left( 3 \right)\left( {2K} \right) \\
& = & \left( 3 \right)\left( {K + K} \right) \\
& > & \left( 3 \right)\left( {K + 1} \right) \\
\end{array}\)

So we can go from the Kth place to the (K+1)th place.

 

[soroban]

Re: Mathematical induction: Prove that 2^n > 3n for all n

Hello, owenthakkar!

Here's my approach . . .

Prove that \(\displaystyle 2^n\:>\:3n\) for all positive integers where \(\displaystyle n\,\geq\,4\)

Note: for \(\displaystyle n\,\geq\,4,\;2^n\:>\: 3\)


First, verify \(\displaystyle S(1).\)


Then assume \(\displaystyle S(k):\;\;2^k \;>\; 3k\)

\(\displaystyle \;\;\)Add \(\displaystyle 2^k\) to both sides: \(\displaystyle \:2^k\,+\,2^k\;> \;3k\,+\,2^k\)

The left side is: \(\displaystyle \,2^k\,+\,2^k\;=\;2\cdot2^k\;=\;2^{k+1}\)

\(\displaystyle \;\;\)So we have: \(\displaystyle \,2^{k+1}\;>\;3k\,+\,2^k\)


Since \(\displaystyle 2^k\,>\,3\), we have: \(\displaystyle \,2^{k+1}\;>\;3k\,+\,2^k\;>\;3k\,+\,3\)

\(\displaystyle \;\;\)Therefore: \(\displaystyle \,2^{k+1}\;>\:3(k\,+\,1)\)