An algebra problem by Anish Roy

Algebra Level 2

The sequence { x n } \{x_{n}\} is defined by x 1 = 1 2 x_{1}=\dfrac{1}{2} and x k + 1 = x k 2 + x k x_{k+1}=x_{k}^{2}+x_{k} for k ≥ 2 k \ge 2 . Find the greatest integer less than 1 x 1 + 1 + 1 x 2 + 1 + 1 x 3 + 1 + ⋯ + 1 x 100 + 1 \large \frac{1}{x_{1}+1} + \frac{1}{x_{2}+1}+ \frac{1}{x_{3}+1}+\cdots+\frac{1}{x_{100}+1}


The answer is 1.

Anish Roy by Anish Roy , 25, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Aug 24, 2017

Similar solution with @Anish Roy 's

1 x k + 1 = 1 x k 2 + x k = 1 x k ( x k + 1 ) = 1 x k − 1 x k + 1 ⟹ 1 x k + 1 = 1 x k − 1 x k + 1 ∑ k = 1 n 1 x k + 1 = ∑ k = 1 n 1 x k − ∑ k = 1 n 1 x k + 1 Note that ∑ k = 1 n 1 x k + 1 = 1 x 1 + 1 + 1 x 2 + 1 + 1 x 3 + 1 + ⋯ + 1 x n + 1 = 1 x 1 − 1 x n + 1 = 2 − 1 x n + 1 < 2 \begin{aligned} \frac 1{x_{k+1}} & = \frac 1{x_k^2 + x_k} \\ & = \frac 1{x_k(x_k+1)} \\ & = \frac 1{x_k} - \frac 1{x_k+1} \\ \implies \frac 1{x_k+1} & = \frac 1{x_k} - \frac 1{x_{k+1}} \\ \color{#3D99F6} \sum_{k=1}^n \frac 1{x_k+1} & = \sum_{k=1}^n \frac 1{x_k} - \sum_{k=1}^n \frac 1{x_{k+1}} & \small \color{#3D99F6} \text{Note that } \sum_{k=1}^n \frac 1{x_k+1} = \frac 1{x_1+1} + \frac 1{x_2+1} + \frac 1{x_3+1} + \cdots + \frac 1{x_n+1} \\ & = \frac 1{x_1} - \frac 1{x_{n+1}} \\ & = 2 - \frac 1{x_{n+1}} < 2 \end{aligned}

⟹ ⌊ ∑ k = 1 n 1 x k + 1 ⌋ = 1 \implies \displaystyle \left \lfloor \sum_{k=1}^n \frac 1{x_k+1} \right \rfloor = \boxed{1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Anish Roy
Aug 19, 2017

We will use the recurrence relation of the sequence to telecope the sum. Since x k + 1 = x k 2 + x k x_{k+1}=x_{k}^{2}+x_{k} , we obtain 1 x k + 1 = 1 x k ( x k + 1 ) = 1 x k − 1 x k + 1 \frac{1}{x_{k+1}}=\frac{1}{x_{k}(x_{k}+1)}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1} ⇒ 1 x k + 1 = 1 x k − 1 x k + 1 \Rightarrow \frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}} ⇒ 1 x 1 + 1 + 1 x 2 + 1 + ⋯ + 1 x 100 + 1 = 1 x 1 − 1 x 101 \Rightarrow \frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{100}+1}=\frac{1}{x_{1}}-\frac{1}{x_{101}} Since 1 x 1 = 2 \frac{1}{x_{1}}=2 and 0 < 1 x 101 < 1 0<\frac{1}{x_{101}}<1 , the integer part of the sum is 1 \boxed{1} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...