Let I (n)=2cos n x, nϵN, then I(1)I(n+1)-I(n)=___
I(n+4)
I(n+2)
I(n+3)
I(1)
I(n)=2 cos n x. I(1)I(n+1)-I(n)=2 cosx. 2cos(n+1)x-2cos n x =2[2 cos (n+1)x cos x-cosnx] =2[cos (n+2)x+cos n x-cos nx] =2 cos (n+2)x =I(n+2)