Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 6, 2011"
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==Solution== | ==Solution== | ||
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<math> 1+11+\cdots+\underbrace{111\cdots 111}_{n}} = n + 10(n-1) + 100(n-2) + \cdots + 10^{n-1}(n-(n-1)</math>. This can be expressed as the summation: | <math> 1+11+\cdots+\underbrace{111\cdots 111}_{n}} = n + 10(n-1) + 100(n-2) + \cdots + 10^{n-1}(n-(n-1)</math>. This can be expressed as the summation: | ||
− | <math> | + | <math>\sum\limits_{k=0}^n 10^{k}(n-k) = n \cdot \frac{10^{n+1}-1}{9} -\sum_{k=0}^n k10^k</math>. Let <math>S = \sum_{k=0}^n k10^k</math>, so that <math>10S = \sum_{k=0}^n (k+1)10^{k+1} - 10^{k+1} = S + (n+1) \cdot 10^{n+1} - \frac{10}{9}\cdot (10^{n+1}-1)</math>. Thus, the answer is <math>n \cdot \frac{10^{n+1}-1}{9} - \frac{(n+1) 10^{n+1}}{9} + \frac{10}{81}(10^{n+1}-1) = \boxed{\frac{10^{n+1} - 9n - 10}{81}}. </math> |
Latest revision as of 20:31, 6 July 2011
Problem
AoPSWiki:Problem of the Day/July 6, 2011
Solution
Trivial Case
:
Solution
$1+11+\cdots+\underbrace{111\cdots 111}_{n}} = n + 10(n-1) + 100(n-2) + \cdots + 10^{n-1}(n-(n-1)$ (Error compiling LaTeX. Unknown error_msg). This can be expressed as the summation:
. Let , so that . Thus, the answer is