If (1+2x+x2)5=∑k=015akxkthen ∑k=07a2k=
Finding the value of ∑k=07a2k:
Given that (1+2x+x2)5=∑k=015akxk
(1+x)10=∑k=015akxk
10C0+10C1x+10C2x2+....+10C10x10=a0+a1x+a2x2+...+a10x10+..+a15x15
Comparing the coefficients of odd terms
a0=10C0,a2=10C2,......a10=10C10,....a12=a14=0
∑k=07a2k=a0+a2+a4+....a14
=1+10C2+10C4+10C6+10C8+10C10+0+0=210-1=29
Hence, the value of ∑k=07a2k= is 29.