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   $x_n<k_n^2<x_{n+1}$
Source of the problem
SMO (senior)-2014 stage 2 problem 4

Topic
Number Theory
Difficulty Level
Medium
Suggested Book
Excursion in Mathematics

Start with hints

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   $x_n<k_n^2<x_{n+1}$  ,  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 ,\( (n+m+1)^2 \in (x_n , x_{n+1}) \)  .    

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron Problem.

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Functional Equation Problem from SMO, 2018 – Question 35

Try this problem from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation. You may use sequential hints if required.

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and the greatest integer.

Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem. You may use sequential hints to solve the problem.

Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015. You may use sequential hints to solve the problem.