Try this beautiful problem from the PRMO II, 2019 based on Missing Integers.
Consider the sequence of numbers [n+\(\sqrt{2n}+\frac{1}{2}\)] for \(n \geq 1\), where [x] denotes the greatest integer not exceeding x. If the missing integers in the sequence are \(n_1<n_2<n_3<...\) then find \(n_{12}\).
Real Numbers
Algebra
Integers
Answer: is 78.
PRMO II, 2019, Question 1
Elementary Algebra by Hall and Knight
\([n+\sqrt{2n}+\frac{1}{2}]\)=[\((\sqrt{n}+\frac{1}{\sqrt{2}})^2\)]
Let P=[\((\sqrt{n}+0.7)^2\)]
given \(n \geq 1\), put n=1 gives P=2
n=2 gives P=4
n=3 gives P=5
n=4 gives P=7
n-5 gives P=8
n=6 gives P=9
n=7 gives P=11
here missing number are
1,3,6,10,... which is following a certain pattern
1, 1+2, 3+3, 6+4, 10+5, 15+6, 21+7, 28+8, 36+9, 45+10, 55+11, 66+12.
so, \(n_{12}\)=78.
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