Factorising a $9$ out of a sum of $9999...$s

Question

Suppose I have the sum

$$S_n = \sum_{k=0}^n a_k(10^k-1).$$

I can factorise a $9$ out of $S_n$ and this gives me

$$S_n = 9(\overbrace{1111111...}^{\text{$n$ times}}a_n + \overbrace{11111...}^{\text{$n-1$ times}}a_{n-1} + \cdots + 11a_1 + a_0).$$

Is it notationally correct to write this as the double sum

$$S_n = 9\sum_{k=0}^n\sum_{j=0}^ka_k10^j.$$

Answer 1

Starting from your claim we have : $$ 9\sum_{k=0}^n \sum_{j=0}^k a_k 10^j = 9\sum_{k=0}^n a_k \sum_{j=0}^k 10^j \overset{(*)}{=} 9\sum_{k=0}^n a_k \frac{10^{k+1} -1}{9} = \sum_{k=0}^n a_k (10^{k+1} -1) $$

$(*)$ Using geometric sum formula.

So it almost matches, just be careful with the index of your second sum. It should go only to $k-1$ and not $k$.

Two remarks :

• To avoid problem with emtpy sum when $k=0$, you can shift the indices of the sequence $(a_i)$
• Also in your first definition of $S_n$ note that $a_0$ will never appear (while it appears after the definition) so you should instead say $\sum_{k=0}^n a_k (10^{k+1} -1)$

Therefore $$ 9\sum_{k=1}^{n+1} \sum_{j=0}^{k-1} a_{k-1} 10^j = \sum_{k=0}^n a_k (10^{k+1} -1) $$

Answer 2

$$S_n=\sum_{k=0}^na_k(10^k-1)=\sum_{k=0}^na_k(10-1)(10^{k-1}+10^{k-2}+\ldots+1)$$

which is what you look for except for the indexing of the second sum.

Answer 3

You give a good argument that it should.

$S_n = \sum_{k=0}^n a_k(10^k-1)$ (Note: if $k = 0; a_k(10^k -1) = a_k*0= 0$)

$S_n = \sum_{k=1}^n a_k(10^k-1)$

$\frac {a^m -1}{a-1} = \sum_{j=0}^{m-1} a^j$

So $S_n = \sum_{k=1}^n a_k(10^k-1)=\sum_{k=1}^n a_k(10-1)\sum_{j=0}^{k-1} 10^j$

$=9\sum_{k=1}^n a_k\sum_{j=0}^{k-1} 10^j=9\sum_{k=1}^n \sum_{j=0}^{k-1} a_k10^j$

You are only off by an index of $1$. The summand is to $k-1$; not $k$.