Generating functions are fun

Calculus Level 4

1 + 3 x + 6 x 2 + 10 x 3 + 15 x 4 + 21 x 5 + ⋯ = 1 1 − A x + A x 2 − B x 3 1 + 3x+ 6x^2 + 10x^3 + 15x^4 + 21x^5 + \cdots = \frac 1{1-Ax+Ax^2-Bx^3}

The equation above holds true for positive integers A A and B B . Find A + B A + B .


The answer is 4.

Vijay Simha by Vijay Simha , 47, USA

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1 solution

Easier solution by Brian Moehring

Consider the following for ∣ x ∣ < 1 |x|<1 , we have:

1 + x + x 2 + x 3 + x 4 + x 5 + ⋯ = 1 1 − x Differentiating both sides w.r.t. x 1 + 2 x + 3 x 2 + 4 x 3 + 5 x 4 + 6 x 5 + ⋯ = 1 ( 1 − x ) 2 Differentiating both sides again 2 + 6 x + 12 x 2 + 20 x 3 + 30 x 4 + 42 x 5 + ⋯ = 2 ( 1 − x ) 3 Dividing both sides by 2 1 + 3 x + 6 x 2 + 10 x 3 + 15 x 4 + 21 x 5 + ⋯ = 1 ( 1 − x ) 3 = 1 1 − 3 x + 3 x 2 − x 3 \begin{aligned} 1+x+x^2+x^3+x^4+x^5+\cdots & = \frac 1{1-x} & \small \color{#3D99F6} \text{Differentiating both sides w.r.t. }x \\ 1+2x+3x^2+4x^3+5x^4+6x^5+ \cdots & = \frac 1{(1-x)^2} & \small \color{#3D99F6} \text{Differentiating both sides again} \\ 2+6x+12 x^2+20x^3+30x^4+42x^5+\cdots & = \frac 2{(1-x)^3} & \small \color{#3D99F6} \text{Dividing both sides by }2 \\ 1+3x+6x^2+10x^3+15x^4+21x^5+\cdots & = \frac 1{(1-x)^3} \\ & = \frac 1{1-3x+3x^2-x^3} \end{aligned}

⟹ A + B = 3 + 1 = 4 \implies A + B = 3+1 = \boxed{4}

My solution:

Consider the following for ∣ x ∣ < 1 |x|<1 , we have:

1 + x + x 2 + x 3 + x 4 + x 5 + ⋯ = 1 1 − x Multiplying both sides by x 2 x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + ⋯ = x 2 1 − x Differentiating both sides w.r.t. x 2 x + 3 x 2 + 4 x 3 + 5 x 4 + 6 x 5 + 7 x 6 + ⋯ = 2 x ( 1 − x ) + x 2 ( 1 − x ) 2 = 2 x − x 2 ( 1 − x ) 2 Differentiating both sides again 2 + 6 x + 12 x 2 + 20 x 3 + 30 x 4 + 42 x 5 + ⋯ = ( 2 − 2 x ) ( 1 − x ) + 4 x − 2 x 2 ( 1 − x ) 3 = 2 ( 1 − x ) 3 Dividing both sides by 2 1 + 3 x + 6 x 2 + 10 x 3 + 15 x 4 + 21 x 5 + ⋯ = 1 ( 1 − x ) 3 = 1 1 − 3 x + 3 x 2 − x 3 \begin{aligned} 1+x+x^2+x^3+x^4+x^5+\cdots & = \frac 1{1-x} & \small \color{#3D99F6} \text{Multiplying both sides by }x^2 \\ x^2+x^3+x^4+x^5 +x^6+x^7+\cdots & = \frac {x^2}{1-x} & \small \color{#3D99F6} \text{Differentiating both sides w.r.t. }x \\ 2x+3x^2+4x^3+5x^4+6x^5+7x^6+ \cdots & = \frac {2x(1-x)+x^2}{(1-x)^2} \\ & = \frac {2x-x^2}{(1-x)^2} & \small \color{#3D99F6} \text{Differentiating both sides again} \\ 2+6x+12 x^2+20x^3+30x^4+42x^5+\cdots & = \frac {(2-2x)(1-x)+4x-2x^2}{(1-x)^3} \\ & = \frac 2{(1-x)^3} & \small \color{#3D99F6} \text{Dividing both sides by }2 \\ 1+3x+6x^2+10x^3+15x^4+21x^5+\cdots & = \frac 1{(1-x)^3} \\ & = \frac 1{1-3x+3x^2-x^3} \end{aligned}

⟹ A + B = 3 + 1 = 4 \implies A + B = 3+1 = \boxed{4}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Just as a note: If you don't multiply by x 2 x^2 in the first step, but otherwise you do every other step the same, it's a little easier.

Brian Moehring - 4 years, 2 months ago

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Thanks, I will submit your solution.

Chew-Seong Cheong - 4 years, 2 months ago

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