The correct option is B 2n−22sinnπ4+2n−2
we have (1+x)n=nC0+nC1x+nC2x2+nC3x3+nC4x4+....
Putting x = - 1;
2n=nC0+nC1+nC2+nC3+nC4.... −−−(1)
Putting x = - 1;
0=nC0−nC1+nC2−nC3+nC4−.... −−−(2)
subtracting (1)from (2); we get
2n−1=nC1+nC3+nC5+nC7+nC9+... −−−−(3)
Putting x = i;
(1+i)n=nC0+nC1i+nC2i2+nC3i3+nC4i4+...⇒2n2(1√2+i1√2)n=nC0+nC1i+nC2(−1)+nC3(−i)+nC4i+...⇒2n2(cosπ4+i sinπ4)n=(nC0−nC2+nC4−...)+i(nC1−nC3++C5−....)⇒2n2eiπn4=(nC0−nC2+nC4−...)+i(nC1−nC3++C5−...)⇒2n2(cosnπ4+i sinnπ4)=(nC0−nC2+nC4−...)+i(nC1−nC3++C5−....)
Comparing the inaginary parts;
⇒2n2sinnπ4=(nC1−nC3++C5−....)−−−−(4)
Adding (3)and (4) ; we get
nC1+nC5+nC9+nC13....=2n−22sinnπ4+2n−2