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Try this beautiful problem from HANOI 2018 based on** Sequence and Series**.

## Sequence and Series – HANOI 2018

Let {\(u_n\)} \(n\geq1\) be given sequence satisfying the conditions \(u_1=0\), \(u_2=1\), \(u_{n+1}=u_{n-1}+2n-1\) for \(n\geq2\). find \(u_{100}+u_{101}\)

- is 13000
- is 10000
- is 840
- cannot be determined from the given information

**Key Concepts**

**S**equence

Series

Number Theory

## Check the Answer

But try the problem first…

Answer: is 10000.

Source

Suggested Reading

HANOI, 2018

Principles of Mathematical Analysis by Rudin

## Try with Hints

First hint

Here \(u_2=1\), \(u_3=3\), \(u_4=6\), \(u_5=10\)

Second Hint

by induction \(u_n=\frac{n(n-1)}{2}\) for every \(n\geq1\)

Final Step

Then \(u_n+u_{n+1}\)=\(\frac{n(n-1)}{2}\)+\(\frac{n(n+1)}{2}\)=\(n^{2}\) for every \(n\geq1\) Then \(u_{100}+u_{101}\)=10000.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s