 # 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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