1.

Prove that `2^n >1+nsqrt(2^(n-1)),AAn >2`where `n`is a positive integer.

Answer» `2^(n) gt 1 + n sqrt(2^(n - 1))`
`implies (2^(n) - 1)/(2 - 1) gt n x 2^((n - 1)//2)`
Now, `(2^(n) - 1)//(2 - 1)` is the sum of a G.P whose first term is 1 and common ratio is 2.
We have to prove that `1 + 2 + 2^(2) + 2^(3) + …. + 2^(n - 1) ge n xx 2^((n - 1)//2)`.
Now,
Using A.M. `ge` G.M., we get
`(1 + 2 + 2^(2) + ..... 2^(n - 1))/(n) ge (1 xx 2 xx 2^(2) xx 2^(3) ... 2^(n - 1))^(1//n)`
Now, `R.H.S = (2^(1 + 2 + 3 +... ( n - 1)))^(1//n)`
`= [2^((n - 1) n//2)]^(1//n) = 2^((n - 1)//2)`
Hence, `1 + 2 + 2^(2) + .... + 2^(n - 1) gt n xx 2^((n - 1)//2)`


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