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Question

If A=[3411], then prove that An=[1+2n4nn12n], where n is any positive integer.

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Solution

We are required to prove that for all nN.

P(n)=An=[1+2n4nn12n]
Let n =1
P(1):A1=[1+2(1)4(1)112(1)]=[3411].........(i)
which is true for n=1.
Let the result be true for n=k.
P(k)=Ak=[1+2k4kk12k].....(ii)
Let n =k+1
Then, P(k+1):Ak+1=[1+2(k+1)4(k+1)k+112(k+1)]=[2k+34k+4k+12k1]
Now, LHS=Ak+1=AkAI=[1+2k4kk12k][3411] [Using Eqs. (i)and (ii)]
=[(1+2k).3+(4k).1(1+2k).(4)+(4k).(1)k.3+(12k).1k.(4)+(12k).(1)]=[3+2k44kk+112k]=[1+2(k+1)4(k+1)k+112(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 nN.


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