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March 17, 2020

Sequence and Series | HANOI 2018

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


Sequence

Series

Number Theory

Check the Answer


Answer: is 10000.

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.

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