If A=[3−41−1], then prove that An=[1+2n−4nn1−2n], where n is any positive integer.
We are required to prove that for all n∈N.
P(n)=An=[1+2n−4nn1−2n]
Let n =1
P(1):A1=[1+2(1)−4(1)11−2(1)]=[3−41−1].........(i)
∴ which is true for n=1.
Let the result be true for n=k.
P(k)=Ak=[1+2k−4kk1−2k].....(ii)
Let n =k+1
Then, P(k+1):Ak+1=[1+2(k+1)−4(k+1)k+11−2(k+1)]=[2k+3−4k+4k+1−2k−1]
Now, LHS=Ak+1=AkAI=[1+2k−4kk1−2k][3−41−1] [Using Eqs. (i)and (ii)]
=[(1+2k).3+(−4k).1(1+2k).(−4)+(−4k).(−1)k.3+(1−2k).1k.(−4)+(1−2k).(−1)]=[3+2k−4−4kk+1−1−2k]=[1+2(k+1)−4(k+1)k+11−2(k+1)]
Therefore, the result is true for n=k+1 whenever it is true for n=k. So, by principle of mathematical induction, it is true for all n∈N.