Series

[xc630]

Hello I need a lot of help. I have to prove that the series:

9+90+900+...+9x10^(n-1) = 10^(n)-1

Any help would be appreciated greatly.

 

[soroban]

Hello, xc630!

$\displaystyle \text{Prove: }\,9\,+\,90\,+\,900\,+\,\cdots\,+\,9\cdot10^{n-1}\;=\;10^n\,-\,1$

The sum of a geometric series is: \(\displaystyle \L\,S_n\;=\;a\cdot\frac{r^n\,-\,1}{r\,-\,1}\)
$\displaystyle \;\;$where $\displaystyle a$ is the first term and $\displaystyle r$ is the common ratio.

Our series has: $\displaystyle \,a\,=\,9,\;r\,=\,10$

Therefore: \(\displaystyle \L\,S_n\;=\;9\cdot\frac{10^n\,-\,1}{10\,-\,1}\;=\;9\cdot\frac{10^n\,-\,1}{9}\;=\;10^n\,-\,1\)

 

[stapel]

Try using the geometric-series summation formula. Or are you supposed to use some other formula?

Thank you.

Eliz.

 

[xc630]

I was taught the sum of a geometric series is Sn= T1 (1-R^N)/ (1-R) where T1 is the 1st term, R is the rate, and where N represents the number of terms. so how owuld I use this formula for the problem?

 

[Mrspi]

I was taught the sum of a geometric series is Sn= T1 (1-R^N)/ (1-R) where T1 is the 1st term, R is the rate, and where N represents the number of terms. so how owuld I use this formula for the problem?

Your formula is the same as the one Soroban showed you....except that Soroban used "a" for the first term, and you used T1. Multiply both numerator and denominator of the fraction by -1:

S<SUB>n</SUB> = T1 [-1*(1 - r<SUP>n</SUP>) / -1(1 - r)]

S<SUB>n</SUB> = T1 [(r<SUP>n</SUP> - 1) / (r - 1)]
See? Your formula is really the same as Soroban's.

Now, T1 = 9, and r = 10....substitute, and simplify the expression.