Let f(θ)=1tan9θ(1+tanθ)10+(2+tanθ)10+…+(20+tanθ)10−20tanθ The left hand limit of f(θ) as θ→π2 is
Let x=tanθ. Then
ftan−1x=(1+x)10+(2+x)10+…+(20+x)10−20x10x9
and x→∞ as θ→π−2
∴ limθ→π−2 f(θ)=limx→∞ ftan−1x=limx→∞ (1+x)10+(2+x)10+…+(20+x)10−20x10x9=limx→∞ ∑r=120 (r+x)10−x10x9=limx→∞ ∑r=120 10C0r10+10C1r9x+…+10C9r9x9=∑r=120 limx→∞ 10C0r10+ 10C1r9x8+…+10C9r=∑r=110 10C9r=10∑r=120 r=10×20×212=2100