If A=⎡⎢⎣100010ab−1⎤⎥⎦, then for n ϵ N, An=
A, if n is odd
A, if n is even
I, if n is odd
I, if n is even
A=⎡⎢⎣100010ab−1⎤⎥⎦;A2=⎡⎢⎣100010ab−1⎤⎥⎦⎡⎢⎣100010ab−1⎤⎥⎦⇒A2=⎡⎢⎣100010001⎤⎥⎦=ISince,A2=I, A3=A2.A=IA=AA4=A3.A=A.A=IIn general, An=A, if n is oddAn=I, if n is even
The value of i1+3+5+......+(2n+1) is