Question 297 of 519MediumMCQ
If ∣x∣<1, ∣y∣<1, and x=y, then the sum to infinity of the series (x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+⋯ is: Solutions · 1
Official solution
Each term is x−yxn+1−yn+1, so the sum is x−y1[1−xx2−1−yy2]=(1−x)(1−y)x+y−xy. Answer the question to unlock