The first two decimal digits of the(n+1)th number square root using sqrt of n
Posted on: March 9, 2025
Everything follows a pattern, we just need to observe it.
First let's see some numbers here:
\[
\begin{aligned}
(\sqrt{3} + \sqrt{4})^2 &= 13.9 \\
(\sqrt{4} + \sqrt{5})^2 &= 17.9 \\
(\sqrt{5} + \sqrt{6})^2 &= 21.9 \\
\vdots
\end{aligned}
\]
It is in arithmetic progression with common difference d = 4
\[ a_n = a_1 + (m-1)d \]
I started the sequence with 3 so to map accordingly, m will be n-2.
For this sequence: a_1 = 13.9, d = 4
Okay, what next?
Let's define the series in one equation:
\[
\begin{aligned}
(\sqrt{n+1} + \sqrt{n})^2 &= X \\
2n + 1 + 2\sqrt{n^2+n} &= X
\end{aligned}
\]
Rearranging for the term with the square root:
\[ \sqrt{n^2+n} = \frac{X - 2n - 1}{2} \]
We know what X is, and it follows an arithmetic progression, so let's substitute that:
\[
\begin{aligned}
\sqrt{n^2+n} &= \frac{13.9 + (n-3)(4) - 2n - 1}{2} \\
&= \frac{13.9 + 4n - 12 - 2n - 1}{2} \\
&= \frac{0.9 + 2n}{2} \\
&= n + 0.45
\end{aligned}
\]
After performing calculations with many numbers using this equation, adding an extra 0.04 gives more accuracy, so
I'm adding it too:
\[
\begin{aligned}
\sqrt{n^2+n} &\approx n + 0.49 \\
\sqrt{n(n+1)} &\approx n + 0.49
\end{aligned}
\]
From this, we can derive:
\[ \sqrt{n+1} \approx \sqrt{n} + \frac{0.49}{\sqrt{n}} \]
Let's try with one number:
n = 1234
\[
\begin{aligned}
\sqrt{n+1} &= \sqrt{1234+1} \\
&= \sqrt{1235} \\
&\approx \sqrt{1234} + \frac{0.49}{\sqrt{1234}} \\
&\approx 35.1282 + \frac{0.49}{35.1282} \\
&\approx 35.1282 + 0.0140 \\
&\approx 35.1422
\end{aligned}
\]
Our interest is in the first 2 decimal digits, so: 35.14
What is the actual (n+1) square root value?
\[ \sqrt{1234+1} = \sqrt{1235} = 35.142566782 \]
We got those 2 digits!