Track / Stage 6 / 254 of 400 #1254 of 1964
Problem 1254 National olympiad, first round Difficulty 6.4 Prove it
## Problema 2
Fie mulțimea A = { n + 1 − n n 2 + n ∣ n ∈ N ∗ } A=\left\{\left.\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^{2}+n}} \right\rvert\, n \in \boldsymbol{N}^{*}\right\} A = { n 2 + n n + 1 − n n ∈ N ∗ }
a) Demonstraţi că n + 1 − n n 2 + n = 1 n − 1 n + 1 \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^{2}+n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} n 2 + n n + 1 − n = n 1 − n + 1 1 n ∈ N ∗ n \in \boldsymbol{N}^{*} n ∈ N ∗
b) Să se demonstreze că: 6 − 5 5 2 + 5 + 7 − 6 6 2 + 6 + ⋯ + 2014 − 2013 2013 2 + 2013 = 1 5 − 1 2014 \frac{\sqrt{6}-\sqrt{5}}{\sqrt{5^{2}+5}}+\frac{\sqrt{7}-\sqrt{6}}{\sqrt{6^{2}+6}}+\cdots+\frac{\sqrt{2014}-\sqrt{2013}}{\sqrt{2013^{2}+2013}}=\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{2014}} 5 2 + 5 6 − 5 + 6 2 + 6 7 − 6 + ⋯ + 201 3 2 + 2013 2014 − 2013 = 5 1 − 2014 1
c) Să se demonstreze că există o submulțime B B B A A A B B B 1 315 \frac{1}{\sqrt{315}} 315 1
This one wants a proof. Work it on paper, then read the official solution and mark
yourself. Be honest about it: the record is only any use to you if it is.
I solved it I didn't Skip
Official solution ## Problema 2
a)
1 n − 1 n + 1 = n + 1 n ( n + 1 ) − n n ( n + 1 ) = n + 1 − n n ( n + 1 ) = n + 1 − n n 2 + n \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=\frac{\sqrt{n+1}}{\sqrt{n(n+1)}}-\frac{\sqrt{n}}{\sqrt{n(n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^{2}+n}} n 1 − n + 1 1 = n ( n + 1 ) n + 1 − n ( n + 1 ) n = n ( n + 1 ) n + 1 − n = n 2 + n n + 1 − n
b)
6 − 5 5 2 + 5 + 7 − 6 6 2 + 6 + ⋯ + 2014 − 2013 2013 2 + 2013 = = 1 5 − 1 6 + 1 6 − 1 7 + 1 7 − 1 8 + ⋯ + 1 2013 − 1 2014 = 1 5 − 1 2014
\begin{aligned}
& \frac{\sqrt{6}-\sqrt{5}}{\sqrt{5^{2}+5}}+\frac{\sqrt{7}-\sqrt{6}}{\sqrt{6^{2}+6}}+\cdots+\frac{\sqrt{2014}-\sqrt{2013}}{\sqrt{2013^{2}+2013}}= \\
& =\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{8}}+\cdots+\frac{1}{\sqrt{2013}}-\frac{1}{\sqrt{2014}}=\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{2014}}
\end{aligned}
5 2 + 5 6 − 5 + 6 2 + 6 7 − 6 + ⋯ + 201 3 2 + 2013 2014 − 2013 = = 5 1 − 6 1 + 6 1 − 7 1 + 7 1 − 8 1 + ⋯ + 2013 1 − 2014 1 = 5 1 − 2014 1
(2p)
1 315 = 1 3 35 = = 1 2 35 − 1 6 35 = 1 140 − 1 1260 = = 1 140 − 1 141 + 1 141 − 1 142 + + ⋯ + 1 1259 − 1 1260 = = 141 − 140 140 2 + 140 + 142 − 141 141 2 + 141 + ⋯ + 1260 − 1259 1259 2 + 1259 ⇒ B = { 141 − 140 140 2 + 140 , 142 − 141 141 2 + 141 , … , 1260 − 1259 1259 2 + 1259 }
\begin{aligned}
& \frac{1}{\sqrt{315}}=\frac{1}{3 \sqrt{35}}= \\
& =\frac{1}{2 \sqrt{35}}-\frac{1}{6 \sqrt{35}}=\frac{1}{\sqrt{140}}-\frac{1}{\sqrt{1260}}==\frac{1}{\sqrt{140}}-\frac{1}{\sqrt{141}}+\frac{1}{\sqrt{141}}-\frac{1}{\sqrt{142}}++\cdots+\frac{1}{\sqrt{1259}}-\frac{1}{\sqrt{1260}}= \\
& =\frac{\sqrt{141}-\sqrt{140}}{\sqrt{140^{2}+140}}+\frac{\sqrt{142}-\sqrt{141}}{\sqrt{141^{2}+141}}+\cdots+\frac{\sqrt{1260}-\sqrt{1259}}{\sqrt{1259^{2}+1259}} \\
& \Rightarrow B=\left\{\frac{\sqrt{141}-\sqrt{140}}{\sqrt{140^{2}+140}}, \frac{\sqrt{142}-\sqrt{141}}{\sqrt{141^{2}+141}}, \ldots, \frac{\sqrt{1260}-\sqrt{1259}}{\sqrt{1259^{2}+1259}}\right\}
\end{aligned}
315 1 = 3 35 1 = = 2 35 1 − 6 35 1 = 140 1 − 1260 1 == 140 1 − 141 1 + 141 1 − 142 1 + + ⋯ + 1259 1 − 1260 1 = = 14 0 2 + 140 141 − 140 + 14 1 2 + 141 142 − 141 + ⋯ + 125 9 2 + 1259 1260 − 1259 ⇒ B = { 14 0 2 + 140 141 − 140 , 14 1 2 + 141 142 − 141 , … , 125 9 2 + 1259 1260 − 1259 }
← Previous All problems Next →
Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
Statement and solution reproduced as published; topic, difficulty and ordering added
by this site.