# How to solve $11+192+1993+19994+...+1999999999$

I thought of converting this to the $$\sum$$ notation and then simplifying it further.

On simplifying, it turned out to be a combination of two summation notations.

Here's what I got

$$\sum_{n = 1}^{9} \left[ 10^n + 90 \left( \sum_{a = 0}^{n - 2} 10^a \right) + n \right]$$
(All edits on the summation notation are welcome)

Was this a correct approach to the problem? If yes, how do I simplify it further?

If no, then what is the right approach and why is this not the right approach? Also, in what types of questions is this the right approach?

Thanks

$$20+200+2000+\cdots2000000000-9-8-7-\cdots1=2222222220-45.$$

($$45$$ is the ninth triangular number.)

• In this particular case, using formulas makes it unnecessarily complicated.
– user65203
Commented Apr 21, 2020 at 7:15
• Thank You for the answer Commented Apr 21, 2020 at 7:16

The $$k$$th summand seems to be $$2\cdot 10^k-10+k$$ (i.e., $$20-9$$, $$200-8$$, $$2000-7$$ and so on). Then we arrive at a geometric and an arithmetic sum:

$$\sum_{k=1}^9(2\cdot 10^k -10+k)=2\sum_{k=1}^910^k-90+\sum_{k=1}^9k=2\cdot\frac{10^{10}-10}9-90+\frac{9\cdot10}2$$

Put differently: Add $$9+8+7+6+5+4+3+2+1$$, and the sum greatly simplifies.

• Thanks :) ${}{}{}$ Commented Apr 21, 2020 at 7:15
• Simpler to consider $k$ from $9$ to $1$ instead of $10-k$.
– user65203
Commented Apr 21, 2020 at 7:17