How to do this:
$$\sum\limits_{k=3}^{n}\bigg\lfloor | \frac {-1-\sqrt {1+8k}} {2}|\bigg\rfloor$$ Where $\lfloor.\rfloor$ denotes floor function and $|.|$ denotes absolute value.
Any hints, method, would help me a lot.
Solving: $\sum\limits_{k=3}^{n}\bigg\lfloor | \frac {-1-\sqrt {1+8k}} {2}|\bigg\rfloor$
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$\begingroup$
real-analysis
algebra-precalculus
elementary-number-theory
functions
floor-function
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1Hint: the term increments by $1$ when $k$ is a triangular number. – 2017-01-27
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2The absolute value is easy to deal with as the inside is always negative. – 2017-01-27
1 Answers
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First of all, observe (trivially) that the sum can be rewritten as:
$$\sum\limits_{k=3}^{n}\bigg\lfloor \frac {1+\sqrt {1+8k}} {2}\bigg\rfloor$$
Secondly, @dxiv's comment is a very good hint.
To see why that's so, consider the formula for the $(n-1)$th triangular number $k$. This is given by $\frac 12 (n)(n-1) = k$.
Solving for $n$ using the quadratic formula gives:
$\displaystyle n = \frac{1 + \sqrt{1 + 8k}}{2}$, which is only integral when $k$ is a triangular number as mentioned in the comment.
All that's left is to observe the pattern in the series. It starts:
$3,3,3,4,4,4,4,5,5,5,5,5,6...$
where it should be obvious that there are $m$ occurences of the natural number $m$ for $m \geq 3$.
I think you should be able to proceed from here.
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0please explain; why $\displaystyle n = \frac{1 + \sqrt{1 + 8k}}{2}$ is integral when $k$ is triangular number. – 2017-01-27
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0Because that quadratic was set up as the solution for the index of the triangular number sequence. It can only be integral when the output of the formula is a triangular number. The rest of the time, $n$ is irrational, but its integer portion is "preserved" - this is why the floor function forces that to "repeat" a certain number of times. – 2017-01-27
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0@mathlover $1+8k=(2m+1)^2 ,k=m(m+1)/2 =$triangular number. – 2017-01-28