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What is the sum of: 6 + 66 + 666 + 6666 ............+ 666666666...6 where there are n sixes in the last term?
Age
16 to 18
Challenge level
Problem
What is the sum of: $$6 + 66 + 666 + 6666 + \cdots + 666666666\cdots6$$ where there are $n$ sixes in the last term?
Did you know ... ?
Many functions, including the trigonometric and exponential functions that you meet in school, can be approximated by infinite power series and good approximations can be found using a finite number of terms. If the series is centred at zero then it can be written in the form $\Sigma_{n=0}^\infty a_nx^n$ where the coefficients depend on the derivative of the function at the origin. The infinite geometric series $1 + x + x^2 + \cdots $ which converges for $|x| < 1$ is the power series for the function $(1 - x )^{-1}$.
Many functions, including the trigonometric and exponential functions that you meet in school, can be approximated by infinite power series and good approximations can be found using a finite number of terms. If the series is centred at zero then it can be written in the form $\Sigma_{n=0}^\infty a_nx^n$ where the coefficients depend on the derivative of the function at the origin. The infinite geometric series $1 + x + x^2 + \cdots $ which converges for $|x| < 1$ is the power series for the function $(1 - x )^{-1}$.
Getting Started
What can you say about the number $111111$?
Can you write $666666$ as a series?
Can you write $666666$ as a series?
Student Solutions
First consider $S_n = 1 + 11 + 111 + 1111 + \cdots$ to $n$ terms. Each individual term can be written and summed as a geometric series, for example $$1111 = 1 + 10 + 100 + 1000 = \frac{10^4-1}{10 - 1}$$ Hence $$S_n= \frac{10^1 - 1}{9} + \frac{10^2 - 1}{9} + \frac{10^3 - 1}{9} + \frac{10^4 - 1}{9} + ... + \frac{10^n - 1}{9}$$ $$= \frac{10 + 10^2 + 10^3 + 10^4 + ... +10^n }{9} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10}{81} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10 - 9n}{81}$$ So $6 + 66 + 666 + 6666 \cdots$ to $n$ terms is: $$6( 1 + 11 + 111 + 1111 + ... ) = \frac{2}{3}\Big[ \frac{10(10^n - 1)}{9}- n \Big]$$