Binomial Theorem

Question

If (1+ax+bx²)⁴=1+8x+32x²+...., find a and b.

Answer 1

Let's rewrite the problem first

(1 + ax + bx²)⁴ = (1+x(a+bx))⁴

= 1 + (₄C₁)[x(a+bx)] + (₄C₂)[x(a+bx)]² + (₄C₃)[x(a+bx)]³ + (₄C₄)[x(a+bx)]⁴

= 1 + 4[x(a+bx)] + 6[x(a+bx)]² + 4[x(a+bx)]³ +[x(a+bx)]⁴

= 1 + 4ax + 4bx² + 6a²x² + 12abx³ + 6b²x⁴ + 4(ax ⁺ bx²)^{3 +} (ax + bx²)⁴

 

Now the only x term is the 4ax

4ax = 8x

and a = 2

 

The x² terms are 4bx² + 6a²x²

4bx² + 6a²x² = 32x²

Substitute the value for a determined above

4bx² + 24x² = 32x²

4bx² = 8x²

and b=2

 

Let's see if this works

(1+2x+2x²)⁴

(1+2x+2x²+2x+4x²+4x³+2x²+4x³+4x⁴)²

(1+4x+8x²+8x³+4x⁴)²

1+4x+8x²+8x³+4x⁴+4x+16x²+32x³+32x⁴+16x⁵+8x²+32x³+64x⁴+64x⁵+32x⁶+8x³+.....

1+8x+32x²+...