0
$\begingroup$

I apologize for forgetting all my high school math!

How can I get the reverse of a summation?

$$x = \frac{n ( n + 1 )}{2}$$

I know that with basic algebra I can do:

$$\sqrt{2x - n} = n$$

But obviously, I need the $n$ on the other side :(

  • 0
    Can you explain what you mean by "reverse of a summation"? Do you mean, given $S_n=\sum_{k=1}^n a_k$, find $a_k$ in terms of $S_n$s? – 2017-02-20
  • 0
    Based on the answer below, what you're asking for should not be called "reverse of a summation." It should be called "solving $x=n(n+1)/2$ for $n$." – 2017-02-20

2 Answers 2

1

I believe you are trying to solve for $n$ while knowing $x=\frac{n(n+1)}{2}$. If so, then you are solving quadratic equation $n^2+n-2x=0$. This equation has two solutions $\frac{1}{2}(-1-\sqrt{1+8x})$ and $\frac{1}{2}(-1+\sqrt{1+8x})$.

  • 0
    Yeah, that's it. Thanks! – 2017-02-20
0

For a quick and dirty way to find $n$ given $x$:

Multiply $x$ by 2, take the square root, and round down to the nearest number.

This always works, since $2x = n(n+1)$, so the square root will always be between $n$ and $n+1$.