# Is $\sum\limits_{{\rm{i}} = 1}^{n - k} {{b_i}} \times {b_{n-2}} + b_{k}$ correct to express this pattern in base n?

having seen the pattern be,ow i have tried to express it in base n, of course there should be few constraints added to parameters.

\begin{align} 1 \times 8 + 1 &= 9\\ 12 \times 8 + 2 &= 98\\ 123 \times 8 + 3 &= 987\\ 1234 \times 8 + 4 &= 9876\\ 12345 \times 8 + 5 &= 98765\\ 123456 \times 8 + 6 &= 987654\\ 1234567 \times 8 + 7 &= 9876543\\ 12345678 \times 8 + 8 &= 98765432\\ 123456789 \times 8 + 9 &= 987654321\\ &\;\vdots\\ \sum\limits_{{\rm{i}} = 1}^{n - k} {{b_i}} \times {b_{n-2}} + b_{k} &= {b_{n - 1}}{b_{n - 2}} \cdot \cdot \cdot {b_{n - k}}\\ \end{align}

where $b_k$ is the kth symbol in base n.

my question is, is that correct? and what are some other ways that the above pattern can be expressed/generalised?

• The way you have expressed it, these are not true. I think the first one is $+1$, second one is $+2$ and so on & then it kind of reverses. Please correct the question – Shailesh Aug 4 '15 at 3:15
• I don't know what you mean, can you please edit the question so I can see what you mean please? – Arjang Aug 4 '15 at 3:25
• Look at the second line... $12$ times $8$ is $96$. Add $1$ to it and you get $97$, and not $98$ as shown in your equation. Also when your doing times 6, it should be $123456$ x $8$ + $6$ = $987654$ and so on further – Shailesh Aug 4 '15 at 3:28
• @Shailesh : thank you, fixed it now, I think – Arjang Aug 4 '15 at 3:33
• It is still not fixed. Please see the RHS. The digits should be in descending order. – Shailesh Aug 4 '15 at 3:54