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# Understand the problem

For each positive integer $n$ let $$x_n=p_1+\cdots+p_n$$ where $p_1,\ldots,p_n$ are the first $n$ primes. Prove that for each positive integer $n$, there is an integer $k_n$ such that $x_n
##### Source of the problem
SMO (senior)-2014 stage 2 problem 4

Number Theory
Medium
##### Suggested Book
Excursion in Mathematics

Do you really need a hint? Try it first!

We have $$P_1 = 2 , P_=3 , P_3=5 , P_4 =7 , P_5 = 11 \ and \ so \ on ….$$. Now to understand the expression $x_n , observe . $$For \ n=1 \ , \ 2 < 2^2 < 2+3$$ $$For \ n=2 \ , \ 2+3 < 3^2 < 2+3+5$$ $$For \ n=3 \ , \ 2+3+5 < 4^2 < 2+3 +5+7$$ $$For \ n=4 \ , \ 2+3 +5+7 < 5^2 <2+3 +5+7 +11$$ Now proceed to prove $$\forall n \geq 5$$ .
Observe $$\forall n \geq 5$$ we have $$P_n > (2n-1)$$. [where $$n \in Z^+$$] Then try to use $$x_n = P_1 + P_2 + …+P_5+….. +P_n > 1 +3 + ….+ 9 +… (2n-1) = n^2 \\ \Rightarrow x_n > n^2 , \forall n \geq 5[where \ n \in Z^+]$$ .
Think if $$x_n=P_1 + P_2 + …. + P_5 +…P_n = b^2 for \ some \ n , b \in Z^+$$ , then we are done . If not so , then think $$m$$ be the largest non negative integer such that $$(n+m)^2 < x_n$$ . Now note that the next perfect square is $$(n+m+1)^2$$ . Observe that if we can prove that $$(n+m+1)^2 – (n+m)^2 = (2n+ 2m +1) \geq P_{n+1}$$ , then we are done . Now try to verify this claim .

Suppose our claim is not true i.e. $$P_n < 2n + 2m +1$$ . So, $$P_n < 2n + 2m +1 \\ \Rightarrow 2n+ 2m \geq P_n , \forall n \in Z^+ \\ \Rightarrow (2n +2m-2)+(2n+ 2m -4)+…..2m \geq P_n + P_{n-1}+……+P_1 \\ \Rightarrow n^2 + 2mn -n \geq P_n + P_{n-1}+……+P_1 \\ \Rightarrow n^2 + 2mn -n \geq x_n \\ \Rightarrow n^2 + 2mn +m^2 > n^2 + 2mn -n\geq x_n \\ \Rightarrow (n+m)^2 > x_n$$ . Contradiction! since we have assumed $$x_n = P_1 + P_2+……+P_{n-1} > (n+m)^2$$ . Thus ,$$(n+m+1)^2 \in (x_n , x_{n+1})$$ .

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