Let (1+x−2x2)20=∑40r=0arxr. then a1+a3+a5+....+a39 is equal to:
-219
-220
-221
-218
Let f(x)=(1+x−2x2)20</br>
Then a1+a3+a5+.......+a39=f(1)−f(−1)2=0−2202=−219
If (1+x+2x2)20=a0+a1+a2x2+...+a40x40, then a1+a3+a5+...+a37 equals :