The constant term of (x+x−32)15 when expanded is 5005.
When the above expression is expanded, we get
Applying Binomial Therom:
∑15i=0(15i)x(15−i)(x−32)i
x15+15x252+105x10+455x152+1365x5+3003x52+5005+6435x52+6435x5+5005x152+3003x10+1365x252+455x15+105x352+15x20+1x452
The constant term as seen above is 5005.
-Vinculum
The constant term of (x+x−32)15 when expanded is 5005.
When the above expression is expanded, we get
Applying Binomial Therom:
∑15i=0(15i)x(15−i)(x−32)i
x15+15x252+105x10+455x152+1365x5+3003x52+5005+6435x52+6435x5+5005x152+3003x10+1365x252+455x15+105x352+15x20+1x452
The constant term as seen above is 5005.
-Vinculum