If n- 0- prove that 2n- 1-n-2n-1

Using the relation A.M. ≥ G.M. on the numbers 1, 2, 2^2, 2^3, ..., 2^n 1, we have 1+2+2^2+...+2^n 1/n gt; ( 1.2.2^2.2^3.....2^n 1 )^1/n Equalit ...

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Standard XII

Mathematics

Arithmetic Mean

If n> 0, pr...

Question

If $n > 0,$ prove that $2^{n} > 1 + n \sqrt{2^{n - 1}}$

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Solution

Using the relation A.M. $\geq$ G.M. on the numbers $1, 2, 2^{2}, 2^{3}, . . ., 2^{n - 1},$ we have
$\frac{1 + 2 + 2^{2} + . . . + 2^{n - 1}}{n} > {({1.2.2}^{2} {.2}^{3} . . . . {.2}^{n - 1})}^{1 / n}$
Equality does not hold as all the numbers are not equal.
$\Rightarrow$ $\frac{2^{n} - 1}{2 - 1} > n {(2^{\frac{(n - 1) n}{2}})}^{\frac{1}{n}}$
$\Rightarrow$ $2^{n} - 1 > n {.2}^{\frac{(n - 1)}{2}}$
$\Rightarrow$ $2^{n} > 1 + n {.2}^{\frac{(n - 1)}{2}}$

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If n>0, prove that 2n>1+n√2n−1

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