If (1-x+x2)n =a0 +a1x +...+ a2nx2n , then a0 + a2 +a4 +....+ a2n=

Question

If (1-x+x2)n =a0 +a1x +...+ a2nx2n , then a0 + a2 +a4 +....+ a2n= (A)  (3n -1/2) (B)  (3n + 1/2) (C)  (2⋅ 3n -1/2) (D)  (2⋅ 3n + 1/2)

Tardigrade Admin · Accepted Answer

(1-x+x2)n =a0 +a1x +a2x2 + ...+ a2nx2n ...(i) Putting x = 1 in (i), we get 1 = a0 +a1 +a2+ .... +a2n ....(ii) Putting x = -1 in (i), we get 3n = a0 +a1 +a2 - a3 +.... +a2n ....(iii) Adding (ii) and (iii), we get 1+3n =2a0 +2a2 +2a4 +...+2a2n ⇒ a0+a2+a4+....a2n =((3n+1/2))

Q. If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + ... + a_{2 n} x^{2 n},$ then $a_{0} + a_{2} + a_{4} + .... + a_{2 n} =$

$\frac{3 ^{n} - 1}{2}$

$\frac{3 ^{n} + 1}{2}$

$\frac{2 \cdot 3 ^{n} - 1}{2}$

$\frac{2 \cdot 3 ^{n} + 1}{2}$

Q. If (1−x+x2)n=a0​+a1​x+...+a2n​x2n, then a0​+a2​+a4​+....+a2n​=

23n−1​

23n+1​

22⋅3n−1​

22⋅3n+1​

Q. If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + ... + a_{2 n} x^{2 n},$ then $a_{0} + a_{2} + a_{4} + .... + a_{2 n} =$

$\frac{3 ^{n} - 1}{2}$

$\frac{3 ^{n} + 1}{2}$

$\frac{2 \cdot 3 ^{n} - 1}{2}$

$\frac{2 \cdot 3 ^{n} + 1}{2}$