Answer : (d) \(\frac{3^n +1}{2}\)
Given, (1 – x + x2)n = a0 + a1x + a2x2 + ..... + a2n x2n ...(i)
Putting x = 1 in the given equation (i), we have
1 = a0 + a1 + a2 + a3 + ..... + a2n ...(ii)
Putting x = –1 in the above given equation (i), we have
(1 – (–1) + 1)n = a0 – a1 + a2 – a3 + ..... + a2n
⇒ 3n = a0 – a1 + a2 – a3 + ..... + a2n ...(iii)
Adding eqns (ii) and (iii), we have
1 + 3n = 2(a0 + a2 + a4 + ..... + a2n)
⇒ a0 + a2 + a4 + ..... + a2n
= \(\frac{3^n +1}{2}\)