Inspired by Sai Ram

Algebra Level 4

The sum of the sequence 0.1 , 0.11 , 0.111 , 0.1111 , … 0.1 , 0.11 , 0.111 , 0.1111 , \ldots upto n n terms is given by e a ( n − 1 b ( 1 − ( 1 c ) d n ) ) \large \frac{e}{a} ( n-\frac{1}{b}(1-(\frac{1}{c})^{dn} )) for positive integers a , b , c , d a,b,c,d and e e with gcd ⁡ ( a , e ) = 1 \gcd(a,e) = 1 .

Find the value of a + b + c + d + e a+b+c+d+e .


The answer is 30.

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Rama Devi by Rama Devi , 43, India

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1 solution

Rama Devi
Jul 25, 2015

t n = ∑ k = 1 n 1 1 0 k = 1 10 ∑ k = 0 n − 1 1 1 0 k = 1 10 ( 1 − ( 1 10 ) n 9 10 ) = 1 9 ( 1 − 1 1 0 n ) t_n = \displaystyle \sum_{k=1}^n \frac{1}{10^k} = \frac{1}{10} \sum_{k=0}^{n-1} \frac{1}{10^k} = \frac{1}{10} \left( \dfrac{1 - \left(\frac{1}{10}\right)^n}{\frac{9}{10}} \right) = \frac{1}{9} \left(1 - \frac{1}{10^n} \right)

∑ k = 1 n t n = 1 9 ∑ k = 1 n ( 1 − 1 1 0 k ) = 1 9 ( n − 1 10 ∑ k = 0 n − 1 1 1 0 k ) = 1 9 ( n − 1 10 ( 1 − 1 1 0 k 9 10 ) ) = 1 9 ( n − 1 9 ( 1 − ( 1 10 ) n ) ) \begin{aligned} \sum_{k=1}^n t_n & = \frac{1}{9} \sum_{k=1}^n \left(1 - \frac{1}{10^k} \right) \\ & = \frac{1}{9} \left(n - \frac{1}{10} \sum_{k=0}^{n-1} \frac{1}{10^k} \right) \\ & = \frac{1}{9} \left(n - \frac{1}{10} \left(\frac{1 - \frac{1}{10^k}}{\frac{9}{10}} \right) \right) \\ & = \frac{1}{9} \left(n - \frac{1}{9} \left(1 - \left(\frac{1}{10}\right)^n \right) \right) \end{aligned}

⇒ a + b + c + d + e = 9 + 9 + 10 + 1 + 1 = 30 \Rightarrow a + b + c + d + e = 9 + 9 +10 + 1 + 1 = \boxed{30}

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Very similar to Chew Seong Sir's solution to a similar problem.

Vishwak Srinivasan - 5 years, 10 months ago

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Yes I know that

Rama Devi - 5 years, 10 months ago

Why don't you upvote

Rama Devi - 5 years, 10 months ago

What happens ?

Sai Ram - 5 years, 10 months ago

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