If sin−1x+sin−1y+sin−1z=3π2, then ∑k=12 x100k+y106k∑x207y207 is
13
43
23
none of these
We have, sin−1x+sin−1y+sin−1z=3π2
It is possible only when
sin−1x=π2,sin−1y=π2 and sin−1z=π2⇒x=1,y=1 and z=1
∴∑k=12 x100k+y106k∑x207y207=x100+y106+x200+y212x207y207+y207z207+z207x207 =1+1+1+11+1+1=43