Question 212 of 298MediumMCQ
If 0<a,b<1 and tan−1a+tan−1b=4π, then the value of (a+b)−2a2+b2+3a3+b3−4a4+b4+⋯ is: Solutions · 1
Official solution
From tan−1a+tan−1b=4π: 1−aba+b=1⟹a+b=1−ab⟹(1+a)(1+b)=2.
The given series is (a−2a2+3a3−⋯)+(b−2b2+3b3−⋯)=ln(1+a)+ln(1+b)=ln[(1+a)(1+b)]=ln2 Answer the question to unlock