11 votes 11 votes Prove using mathematical induction for $n \geq 5, 2^n > n^2$ Set Theory & Algebra gate1995 set-theory&algebra proof mathematical-induction descriptive + – Kathleen asked Oct 8, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 1.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 13 votes 13 votes Base case: $n = 1, 2^1 = 2 > 1^2$ Induction hypothesis: $2^n > n^2$ To prove: $2^{n+1} > {(n+1)}^2$ $\text{LHS} = 2. 2^n > 2. n^2 $ (Induction hypothesis) $\text{RHS} = {(n+1)}^2 = n^2 + 2n + 1 < \text{LHS}$, hence proved. Arjun answered May 31, 2015 edited Mar 24, 2021 by soujanyareddy13 Arjun comment Share Follow See all 2 Comments See all 2 2 Comments reply Warrior commented Aug 6, 2017 reply Follow Share @ Arjun sir in this qs Basis should be n=5 because at n=2,n=3,n=4 eqn 2^n > n^2 does not hold. 6 votes 6 votes vishnu_m7 commented Jun 15, 2020 reply Follow Share Shouldn't the base case be n=5? 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes http://math.stackexchange.com/questions/497092/proof-by-induction-2n-n2-for-all-integer-n-greater-than-4 Anu answered Jun 28, 2015 Anu comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Putting the values of n = 1,2,3,4,5,6,7,8 we find 2^>1^2 2^2=2^2 2^3 <3^2 2^4= 4^2 2^ 5> 5^2 2^6 >6 ^2 2^7 > 7 ^2 and so on Hence we conclude that the given statement holds good for values n> 5 ( by induction) DIBAKAR MAJEE answered Apr 26, 2020 DIBAKAR MAJEE comment Share Follow See all 0 reply Please log in or register to add a comment.