Inspired by Sai Ram

Algebra Level 4

The sum of the sequence 0.6 , 0.66 , 0.666 , 0.6666 , … 0.6 , 0.66 , 0.666 , 0.6666 , \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 25.

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Rama Devi by Rama Devi , 43, India

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1 solution

Chew-Seong Cheong
Jul 22, 2015

We note that the term of the series is:

t n = ∑ k = 1 n 6 1 0 k = 6 10 ∑ k = 0 n − 1 1 1 0 k = 3 5 ( 1 − ( 1 10 ) n 9 10 ) = 2 3 ( 1 − 1 1 0 n ) t_n = \displaystyle \sum_{k=1}^n \frac{6}{10^k} = \frac{6}{10} \sum_{k=0}^{n-1} \frac{1}{10^k} = \frac{3}{5} \left( \dfrac{1 - \left(\frac{1}{10}\right)^n}{\frac{9}{10}} \right) = \frac{2}{3} \left(1 - \frac{1}{10^n} \right)

∑ k = 1 n t n = 2 3 ∑ k = 1 n ( 1 − 1 1 0 k ) = 2 3 ( n − 1 10 ∑ k = 0 n − 1 1 1 0 k ) = 2 3 ( n − 1 10 ( 1 − 1 1 0 k 9 10 ) ) = 2 3 ( n − 1 9 ( 1 − ( 1 10 ) n ) ) \begin{aligned} \sum_{k=1}^n t_n & = \frac{2}{3} \sum_{k=1}^n \left(1 - \frac{1}{10^k} \right) \\ & = \frac{2}{3} \left(n - \frac{1}{10} \sum_{k=0}^{n-1} \frac{1}{10^k} \right) \\ & = \frac{2}{3} \left(n - \frac{1}{10} \left(\frac{1 - \frac{1}{10^k}}{\frac{9}{10}} \right) \right) \\ & = \frac{2}{3} \left(n - \frac{1}{9} \left(1 - \left(\frac{1}{10}\right)^n \right) \right) \end{aligned}

⇒ a + b + c + d + e = 3 + 9 + 10 + 1 + 2 = 25 \Rightarrow a + b + c + d + e = 3 + 9 +10 + 1 + 2 = \boxed{25}

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An excellent solution.

Sai Ram - 5 years, 10 months ago

Mine was a similar solution . it is exactly the same

Sai Ram - 5 years, 10 months ago

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Ya, it is an excellent solution because it is same as yours. Just joking.

Chew-Seong Cheong - 5 years, 10 months ago

Sir I got my answer right and then did a summing mistake and kept entering 24,23,22.

Department 8 - 5 years, 9 months ago

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