Answer to Question #86494 in Algebra for RAKESH DEY

Question

Prove that 2^n>1+ n*√(2^(n-1)) for all n>2 using the inequalities.

Answer 1

Consider a function and find its derivative:

$$f'(n)=(2^{n}-1-n*\sqrt{2^{n-1}})$$

Differentiate the sum term by term and factor out constants:

$$d(-1)/dn+d(2^n)/dn-d(\sqrt{2^{n-1}}*n)/dn$$

The derivative of -1 is zero:

$$d(2^n)/dn-d(\sqrt{2^{n-1}}*n)/dn+0$$

The derivative of 2^n is 2^nlog(2):

$$-(d(\sqrt{2^{n-1}}*n)/dn)+2^nlog(2)$$

Use the product rule, $d(u v)/dn=(v)du/dn+(u)dv/dn$ where $u=\sqrt{2^{n-1}}$ and $v=n$

Simplify the expression:

$$2^nlog(2)-n(d(\sqrt{2^{n-1}})/dn)-\sqrt{2^{n-1}}*(d(n)/dn)$$

Similarly, we repeat the decomposition and as a result we get:

$$-\sqrt{2^{n-1}}+2^n*log(2)-(\sqrt{2^{n-1}}*n*log(2))/2$$

Simplify the expression:

$$log(2)(2^n-n\sqrt{2^{n-1}})-\sqrt{2^{n-1}}$$

As you can see, the derivative is positive when n>2, therefore, the function is increasing at n>2.

Insofar as $f(2)=log(2)(4-\sqrt{2})-\sqrt{2}=0.38$ that function positive for all n>2

Thus, for n> 2, the following inequality holds:$f(n)>0, \implies 2^n-1-n\sqrt{2^{n-1}}>0, \implies$ $2^n>1+n\sqrt{2^{n-1}}$