2

This is probably something stupid and for some reason I can't see it, but let's say you are taking the sum of the first $10^{i}$ squares. If $i=1$ we get $385$,$i=2$ we get $338350$, $i=3$ we get $333833500$ ... the obvious pattern for the digits is $i$ three's, 1 eight, $i-1$ three's, 1 five, and $i-1$ zeros.

I couldn't see why this should emerge, I tried manipulating the standard sum-of-squares formula - $(n/6)(n+1)(2n+1)$ - and just let $n=10^{i}$ but it didn't build any understanding for me.

user115411
  • 465

1 Answers1

2

Note that $$\frac {2\cdot 10^n +1}3= 6\times \frac {10^n-1}9 +1$$

This is of the form $6666\dots 67$ with $n-1$ sixes.

Now note that $\frac {10^n}2=5\cdot 10^{n-1}$. The $10^{n-1}$ part simply adds $n-1$ trailing zeros. Multiply by $5$ therefore.

This gives $$30\times \frac {10^n-1}9 +5$$ which is of the form $333\dots 5$ with $n$ threes.

Finally multiply by $10^n+1$ and see that a $3$ and a $5$ overlap in the middle. So the result will be $$333 \dots 38333 \dots 3500 \dots 00$$

Counting a little more carefully you get $n$ threes followed by one $8$ then $n-1$ threes then $5$ and $n-1$ zeros.

Mark Bennet
  • 100,194