If (1 + x – 3x2)10 = 1 + a1x + a2x2 + ..... + a20x20, then find a2 +

1.	If (1 + x – 3x2)10 = 1 + a1x + a2x2 + ..... + a20x20, then find a2 + a4 + a6 + ..... + a20.
Answer» (1 + x – 3x²)¹⁰ = 1 + a₁x + a₂x² + ..... + a₂₀x²⁰ Putting x = 1 in (i), we get: 1 + a₁ + a₂ + ..... + a₂₀ = (– 1)¹⁰ = 1 Putting x = – 1 in (i), we get 1 – a₁ + a₂ – ..... + a₂₀ = (– 3)¹⁰ = 3¹⁰ Adding (ii) and (iii), we get 2(1 + a₂ + a₄ + ..... + a₂₀) = 3¹⁰ + 1 ⇒ 2(a₂ + a₄ + ..... + a₂₀) = 3¹⁰ – 1 ⇒ a₂ + a₄ + ..... + a₂₀= \(\frac{3^{10} -1}{2}\)

If (1 + x – 3x2)10 = 1 + a1x + a2x2 + ..... + a20x20, then find a2 + a4 + a6 + ..... + a20.

Answer»

(1 + x – 3x²)¹⁰ = 1 + a₁x + a₂x² + ..... + a₂₀x²⁰

Putting x = 1 in (i), we get:

1 + a₁ + a₂ + ..... + a₂₀ = (– 1)¹⁰ = 1

Putting x = – 1 in (i), we get

1 – a₁ + a₂ – ..... + a₂₀ = (– 3)¹⁰ = 3¹⁰

Adding (ii) and (iii), we get

2(1 + a₂ + a₄ + ..... + a₂₀) = 3¹⁰ + 1

⇒ 2(a₂ + a₄ + ..... + a₂₀) = 3¹⁰ – 1

⇒ a₂ + a₄ + ..... + a₂₀= \(\frac{3^{10} -1}{2}\)