Unsquaring a sequence

Algebra Level 4

Let { x n } \{x_n\} be a sequence of positive numbers. If x 1 = 3 2 x_1=\frac{3}{2} and x n + 1 2 − x n 2 = 1 ( n + 2 ) 2 − 1 n 2 x_{n+1}^2-x_{n}^2=\dfrac{1}{(n+2)^2}-\dfrac{1}{n^2} for all positive integer n n .

Find x 1 + x 2 + . . . + x 2009 x_1+x_2+...+x_{2009} .

Challenge: Feel easy? Try out the whole set of problems here .


The answer is 2009.999502.

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Raymond Chan by Raymond Chan , 20, USA

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1 solution

x n + 1 2 − x n 2 = 1 ( n + 2 ) 2 − 1 n 2 Given ∑ k = 1 n ( x k + 1 2 − x k 2 ) = ∑ k = 1 n ( 1 ( k + 2 ) 2 − 1 k 2 ) x n + 1 2 − x 1 2 = 1 ( n + 1 ) 2 + 1 ( n + 2 ) 2 − 1 1 − 1 4 Given that x 1 = 3 2 ⟹ x n + 1 2 = 1 + 1 ( n + 1 ) 2 + 1 ( n + 2 ) 2 Replace n + 1 with n x n 2 = 1 + 1 n 2 + 1 ( n + 1 ) 2 ⟹ x n = x 4 + 2 x 3 + 3 x 2 + 2 x + 1 n 2 ( n + 1 ) 2 = ( n 2 + n + 1 ) 2 n 2 ( n + 1 ) 2 = n 2 + n + 1 n ( n + 1 ) = 1 + 1 n − 1 n + 1 \begin{aligned} x_{n+1}^2 - x_n^2 & = \frac 1{(n+2)^2} - \frac 1{n^2} & \small \color{#3D99F6} \text{Given} \\ \sum_{k=1}^n \left(x_{k+1}^2 - x_k^2\right) & = \sum_{k=1}^n \left(\frac 1{(k+2)^2} - \frac 1{k^2}\right) \\ x_{n+1}^2 - \color{#3D99F6} x_1^2 & = \frac 1{(n+1)^2} + \frac 1{(n+2)^2} - \frac 11 - \frac 14 & \small \color{#3D99F6} \text{Given that }x_1 = \frac 32 \\ \implies x_{n+1}^2 & = 1 + \frac 1{(n+1)^2} + \frac 1{(n+2)^2} & \small \color{#3D99F6} \text{Replace }n+1 \text{ with }n \\ x_n^2 & = 1 + \frac 1{n^2} + \frac 1{(n+1)^2} \\ \implies x_n & = \sqrt{\frac {x^4+2x^3+3x^2+2x+1}{n^2(n+1)^2}} \\ & = \sqrt{\frac {(n^2+n+1)^2}{n^2(n+1)^2}} \\ & = \frac {n^2+n+1}{n(n+1)} \\ & = 1 + \frac 1n - \frac 1{n+1} \end{aligned}

Therefore,

∑ n = 1 2009 x n = ∑ n = 1 2009 ( 1 + 1 n − 1 n + 1 ) = 2009 + 1 − 1 2010 ≈ 2009.999502 \begin{aligned} \sum_{n=1}^{2009} x_n & = \sum_{n=1}^{2009} \left(1 + \frac 1n - \frac 1{n+1}\right) = 2009 + 1 - \frac 1{2010} \approx \boxed{2009.999502} \end{aligned}

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