What kind of a sequence is this?

Calculus Level 4

1 + 8 9 ( 1 0 1 1 ) 1 9 1 + 8 9 ( 1 0 2 1 ) 1 9 2 + 8 9 ( 1 0 3 1 ) 1 9 3 + = ? 1 + \dfrac{\frac89 (10^1 - 1)}{19^1} + \dfrac{\frac89 (10^2 - 1)}{19^2} + \dfrac{\frac89 (10^3 - 1)}{19^3} + \ldots = \ ?

Give your answer to three decimal places.


The answer is 1.93827.

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2 solutions

Alan Yan
Aug 24, 2015

S = 1 + 8 19 + 88 1 9 2 + 888 1 9 3 + . . . S = 1 + \frac{8}{19} + \frac{88}{19^2} + \frac{888}{19^3}+ ...

S 19 = 1 19 + 8 1 9 2 + 88 1 9 3 + . . . \frac{S}{19} = \frac{1}{19}+\frac{8}{19^2}+\frac{88}{19^3} + ...

This implies that you have:

10 S 19 = 10 19 + 80 1 9 2 + 880 1 9 3 + . . . = 10 19 + ( 88 1 9 2 + 888 1 9 3 + . . . ) ( 8 1 9 2 + 8 1 9 3 + . . . ) \frac{10S}{19} = \frac{10}{19}+\frac{80}{19^2}+\frac{880}{19^3} + ... = \frac{10}{19} + (\frac{88}{19^2} + \frac{888}{19^3} + ...) - (\frac{8}{19^2}+\frac{8}{19^3} + ...)

10 S 19 = 10 19 + ( S 27 19 ) 4 171 S = 157 81 1.938 \frac{10S}{19} = \frac{10}{19} + (S - \frac{27}{19}) - \frac{4}{171} \implies S = \frac{157}{81} \approx \boxed{1.938}

You need to check your working again. It should be 157 81 \dfrac{157}{81} .

Pi Han Goh - 5 years, 9 months ago

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oops, that was a mistype.

Alan Yan - 5 years, 9 months ago

@Pi Han Goh Yep the answer is 157 81 \dfrac{157}{81}

Hrishik Mukherjee - 5 years, 9 months ago

The first term is 0, not 1

Jason S - 5 years, 9 months ago
Jp Delavin
Sep 4, 2015

S = 1 + 8 9 ( 1 0 1 1 ) 1 9 1 + 8 9 ( 1 0 2 1 ) 1 9 2 + . . . S = 1 + \frac{\frac{8}{9}(10^1-1)}{19^1} + \frac{\frac{8}{9}(10^2-1)}{19^2} + ...

= 1 + 8 9 n = 0 1 0 n 1 1 9 n = 1 + \frac{8}{9}\sum\limits_{n=0}^\infty \frac{10^n-1}{19^n}

= 1 + 8 9 n = 0 [ 1 0 n 1 9 n 1 1 9 n ] =1 + \frac{8}{9}\sum\limits_{n=0}^\infty \left[ \frac{10^n}{19^n} - \frac{1}{19^n} \right]

= 1 + 8 9 [ 1 1 10 19 1 1 1 19 ] =1 + \frac{8}{9}\left[ \frac{1}{1-\frac{10}{19}} - \frac{1}{1-\frac{1}{19}} \right]

S = 157 81 1.938 S=\frac{157}{81} \approx \boxed{1.938}

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