Prove by mathematical induction that for any integer n Great

Prove by mathematical induction that for any integer n Greaterthanorequalto 4, 2^n Greaterthanorequalto n^2. Prove by induction that the sum of the geometric series (with a not equal to one) is: 1 + a + a^2 + a^n = 1 - a^n + 1/1 - a Prove by induction that the number of different binary strings with n bits is 2^n Prove by mathematical induction that for every positive integer n, sigma_j = 1^n j^3 = (n(n + 1)/2)^2.

Solution

I have solved the first two problems, post one more question to get the remaining two answers

1) Base Case: (n=0 and n=1)

For n=0, LHS = 2^0 = 1, RHS = 0^2 = 0, LHS >= RHS

For n=1.LHS = 2^1 = 2, RHS = 1^2 = 1, LHS >= RHS

Assumption Step (n=k)

Let us assume that the given thing holds for n=k,

2^k >= k^2

Inductive hypothesis step

We need to prove that the given thing holds for n=(k+1)

LHS = 2^(k+1) = 2.2^k = 2k^2

RHS = (k+1)^2 = k^2 + 1 + 2k

LHS >= RHS

k^2 - 1 - 2k >= 0

(k-1)^2 >= 0

since the perfect square is always greater than equal to zero, hence using mathematical induction we can say that

2^n >= n^2

2)

Base Case:(n=0)

LHS = 1

RHS = (1 -a)/(1-a) = 1

Hence the base case is satisfied

Assumption Step (n=k)

Let us assume that the given thing holds for n=k,

1 + a + a^2 + ... + a^k = (1 - a^(k+1))/(1-a)

Inductive hypothesis step

We need to prove that the given thing holds for n=(k+1)

LHS = 1 + a + a^2 + ... + a^k + a^(k+1) = [1 - a^(k+1)]/(1-a) + a^(k+1) = (1 - a^(k+1+1))/(1-a)

Hence the relation is true using mathematical induction