If `A=[costhetaisinthetaisinthetacostheta]` , then prove by principle of mathematical induction that `A^n=[cosn\ thetaisinn\ thetaisinn\ thetacosn\ theta]` for all `n in N` .

If A={:[(cos theta,isintheta),(i sintheta,costheta)]:}, then prove by principle of mathematical induction that A^(n)={:[(cosntheta,isintheta),(isintheta,cosntheta)]:} for all n in N.

If A= [[Cos theta, is in theta],[is in theta cos theta]] , then prove by principle of mathematics induction that: A^n= [[Cos theta, is in theta],[is in theta cos theta]], n in N.

Prove by the principle of mathematical induction that 2^ n >n for all n∈N.

Prove by the principle of mathematical induction that n<2^(n) for alln in N

Prove by the principle of mathematical induction that n<2^n"for all"n in Ndot

Using the principle of mathematical induction, prove that n<2^n for all n in N

Using the principle of mathematical induction, prove that n<2^(n) for all n in N

Prove by the principle of mathematical induction that for all n in N ,n^2+n is even natural number.

Prove by the principle of mathematical induction that for all n in N ,n^2+n is even natural number.

By Principle of Mathematical Induction, prove that : 2^n >n for all n in N.