If 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n , then prove that a 0 + a 1 + a 2 + ⋯ + a 2 n = 3 n .

Given, 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n Put, x = 1 ⇒ ( 1 + 1 + 1 ) n = a 0 + a 1 + a 2 + … + a 2 n ∴ a 0 + a 1 + a 2 + … + a 2 n = 3 n Hence, proved.

If 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n , then prove that a 0 + a 2 + a 4 + ⋯ + a 2 n = 3 n + 1 2 .

Given, 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n ...(i) Put, x = 1 in (i) ⇒ ( 1 + 1 + 1 ) n = a 0 + a 1 + a 2 + … + a 2 n ⇒ a 0 + a 1 + a 2 + … + a 2 n = 3 n ...(ii) Put, x = - 1 in (i), we get ( 1 - 1 + 1 ) n = a 0 - a 1 + a 2 - … + a 2 n ⇒ a 0 - a 1 + a 2 - … + a 2 n = 1 ...(iii) (ii) + (iii) ⇒ 2 a 0 + a 2 + a 4 + … + a 2 n = 3 n + 1 ∴ a 0 + a 2 + a 4 + … + a 2 n = 3 n + 1 2 Hence, proved.

If 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n , then prove that a 1 + a 3 + a 5 + ⋯ + a 2 n - 1 = 3 n - 1 2 .

Given, 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n ...(i) Put, x = 1 in (i), ⇒ ( 1 + 1 + 1 ) n = a 0 + a 1 + a 2 + … + a 2 n ⇒ a 0 + a 1 + a 2 + … + a 2 n = 3 n ...(ii) Put, x = - 1 in (i), we get ( 1 - 1 + 1 ) n = a 0 - a 1 + a 2 - … + a 2 n ⇒ a 0 - a 1 + a 2 - … + a 2 n = 1 ...(iii) (ii) - (iii) ⇒ 2 a 1 + a 3 + a 5 + … + a 2 n - 1 = 3 n - 1 ∴ a 1 + a 3 + a 5 + … + a 2 n - 1 = 3 n - 1 2 Hence, proved.

If 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n , then prove that a 0 + a 3 + a 6 + a 9 + ⋯ = 3 n - 1 .

Given, 1 + x + x 2 n = a 0 + a 1 x + a 2 x 2 + … + a 2 n x 2 n ...(i) Put x = 1 in (i), ⇒ ( 1 + 1 + 1 ) n = a 0 + a 1 + a 2 + … + a 2 n ⇒ a 0 + a 1 + a 2 + … + a 2 n = 3 n ...(ii) We know that, 1 + ω + ω 2 = 0 and ω 3 = 1 Put x = ω in (i), we get a 0 + a 1 ω + a 2 ω 2 + a 3 ω 3 + … + a 2 n ω 2 n = 0 ...(iii) Put x = ω 2 in (i), we get a 0 + a 1 ω 2 + a 2 ω 4 + a 3 ω 6 + … + a 2 n ω 4 n = 0 ...(iv) Adding (ii), (iii) and (iv), we get 3 a 0 + a 1 1 + ω + ω 2 + a 2 1 + ω 2 + ω 4 + a 3 1 + ω 3 + ω 6 + … + a 2 n 1 + ω 2 n + ω 4 n = 3 n ⇒ 3 a 0 + a 1 ( 0 ) + a 2 ( 0 ) + 3 a 3 + … + … = 3 n ⇒ a 0 + a 3 + a 6 + a 9 + … = 3 n 3 ∴ a 0 + a 3 + a 6 + a 9 + … = 3 n - 1 Hence, proved.

If 1 + x + x 2 + x 3 7 = b 0 + b 1 x + b 2 x 2 + … + b 21 x 21 , then find the value of b 0 + b 2 + b 4 + … + b 20 .

Given, 1 + x + x 2 + x 3 7 = b 0 + b 1 x + b 2 x 2 + … + b 21 x 21 ...(i) Put x = 1 in (i), we get b 0 + b 1 + b 2 + … + b 20 + b 21 = 4 7 ...(ii) Put x = - 1 in (i), we get b 0 - b 1 + b 2 + … + b 20 - b 21 = 0 ...(iii) (ii) + (iii) ⇒ 2 b 0 + 2 b 2 + 2 b 4 + … + 2 b 20 = 4 7 ⇒ b 0 + b 2 + b 4 + … + b 20 = 2 2 7 2 ∴ b 0 + b 2 + b 4 + … + b 20 = 2 13

If 1 + x + x 2 + x 3 7 = b 0 + b 1 x + b 2 x 2 + … + b 21 x 21 , then find the value of b 1 + b 3 + b 5 + … + b 21 .

Given, 1 + x + x 2 + x 3 7 = b 0 + b 1 x + b 2 x 2 + … + b 21 x 21 ...(i) Put x = 1 in (i), we get b 0 + b 1 + b 2 + … + b 20 + b 21 = 4 7 ...(ii) Put x = - 1 in (i), we get b 0 - b 1 + b 2 + … + b 20 - b 21 = 0 ...(iii) (ii) - (iii) ⇒ 2 b 1 + 2 b 3 + 2 b 5 + … + 2 b 21 = 4 7 ⇒ b 1 + b 3 + b 5 + … + b 21 = 2 2 7 2 ∴ b 1 + b 3 + b 5 + … + b 21 = 2 13