A=(25−23amp;23amp;−21)
A2=(4−3amp;3amp;−2)
A3=(211−29amp;29amp;−27)
A4=(7−6amp;6amp;−5)
and so on
An=(23n+1−23namp;23namp;−23n+1)
Now
A2022amp;=(23×2022+12−3×2022amp;23×2022amp;2−3×2022+1)amp;=(3034−3033amp;3033amp;−3032)
∴ Option (A) is correct