If an+1=121+an, then cos1-a02a1a2…….∞=
-1
1
a0
1a0
Given an+1=1+an2 Let a0=cosθ⇒a1=1+cosθ2=cosθ2 a2=cosθ4 a3=cosθ8 an+1=cosθ2n+1 and so on
We know that cosθ2cosθ22cosθ23..............cosθ2n−1............∞=sinθθ
∴cos1-a02a1a2…….∞=cossinθsinθ·θ =cosθ=a0