Geometric series

[nanase]

I have a challenging Geometric series question that I am stuck and need help with :

I tried doing some workings but need help in figuring out the ratio, and forming the expression.
PS: I realised the power in the brackets is wrong, it should be equal to n, but I am still stuck otherwise.
Thank you for any help

 

[nanase]

I gave 2 different attempts/approaches above, but I am stuck in how to obtain the expression

 

[topsquark]

I have a challenging Geometric series question that I am stuck and need help with :
View attachment 34140
I tried doing some workings but need help in figuring out the ratio, and forming the expression.
View attachment 34141PS: I realised the power in the brackets is wrong, it should be equal to n, but I am still stuck otherwise.
Thank you for any help

Your second approach is correct, though you didn't apply it correctly. Each term is $6 \cdot (1 + 10 + 100 + \dots )$. So we have
$\displaystyle 6 \cdot \left ( \sum_{k = 1}^1 10^{k - 1} + \sum_{k = 1}^2 10^{k - 1} + \sum_{k = 1}^3 10^{k -1} + \dots \right )$

The problem you were having above is that the ratio in each term is 10. Not, 1, 10, 100, etc.

Can you finish?

-Dan

 

[nanase]

Your second approach is correct, though you didn't apply it correctly. Each term is $6 \cdot (1 + 10 + 100 + \dots )$. So we have
$\displaystyle 6 \cdot \left ( \sum_{k = 1}^1 10^{k - 1} + \sum_{k = 1}^2 10^{k - 1} + \sum_{k = 1}^3 10^{k -1} + \dots \right )$

The problem you were having above is that the ratio in each term is 10. Not, 1, 10, 100, etc.

Can you finish?

-Dan

is this like expression within an expression? I am thinking of including a Series expression for each term, but that is my confusion how can I create an expression if this goes to infinity?

 

[topsquark]

is this like expression within an expression? I am thinking of including a Series expression for each term, but that is my confusion how can I create an expression if this goes to infinity?

6 + 66 + 666 + .. does go to infinity after an infinite number of terms. So you should expect that. But the question is asking about the sum of only n terms.

To write the full expression: The sum of the first n terms is
$\displaystyle 6 + 66 + 666 + \dots = \sum_{j = 1}^n \left ( 6 \sum_{k = 1}^j 10^{k-1} \right )$

However, you need to be able to do each individual sum and that's where you were having a problem. How can you do one of these?

-Dan

 

[blamocur]

6 + 66 + 666 + .. does go to infinity after an infinite number of terms. So you should expect that. But the question is asking about the sum of only n terms.

To write the full expression: The sum of the first n terms is
$\displaystyle 6 + 66 + 666 + \dots = \sum_{j = 1}^n \left ( 6 \sum_{k = 1}^j 10^{k-1} \right )$

However, you need to be able to do each individual sum and that's where you were having a problem. How can you do one of these?

-Dan

Alternately:
$$S_n = 6\left(\sum_{k=1}^{n} \frac{10^k-1}{9} \right)$$

 

[nanase]

I haven't learned Sigma notation so I am having a bit of problem in reading the notation.
I am self learning how to solve questions like this from a book. The answer key is [2(10^(n+1) - 10 - 9n)]/27
I have no means of checking whether that is correct, but I can read it if it is written in that form.
But can somebody help me how to obtain that expression?

 

[nanase]

I did further steps from the help, stuck in the concluding the expression

 

[topsquark]

View attachment 34146
I did further steps from the help, stuck in the concluding the expression

Okay, first put things in standard form. The series $1 + 10 + 100 + \dots$ has a formula for each term $10^{k -1} = \dfrac{1}{10} 10^k$. So $a = \dfrac{1}{10}$ and $r = 10$.

The formula for the sum is $S = \dfrac{a(r^j - 1)}{r - 1}$, where j is the number of terms in the sum What's S?

-Dan

 

[nanase]

Hi Mr Dan,
sorry I got distracted with exam for past few days, but I am still very interested in solving this,
I am stuck in finding the ratio. Is it 10 ? I tried dividing the consecutive terms to find r but it turns out they are not the same?
In your previous post you put 10^k - 1, where is the position of that -1, next to the k? or next to 10^k.
I am dealing with series within series here, but I guess forming the expression is my biggest hurdle..
please help

 

[topsquark]

View attachment 34248Hi Mr Dan,
sorry I got distracted with exam for past few days, but I am still very interested in solving this,
I am stuck in finding the ratio. Is it 10 ? I tried dividing the consecutive terms to find r but it turns out they are not the same?
In your previous post you put 10^k - 1, where is the position of that -1, next to the k? or next to 10^k.
I am dealing with series within series here, but I guess forming the expression is my biggest hurdle..
please help

Well, it would've helped if I had given you the correct formula! Sorry about that.

$6 + 66 + 666 + \dots = 6(1 + 11 + 111 + \dots )$

Each term can be expressed as the sum of a series $1 + 10 + 100 + \dots$ which is a geometric series of j terms with a = 1/10 and r = 10. The sum of each will be
$S_j = \dfrac{a r (r^j - 1)}{r - 1}$ <-- Notice the extra r in there!

So
$6 + 66 + 666 + \dots = 6 \left ( \dfrac{ \dfrac{1}{10} \cdot 10 ( 10^1 - 1 ) }{10 -1} + \dfrac{ \dfrac{1}{10} \cdot 10 ( 10^2 - 1 ) }{10 -1} + \dfrac{ \dfrac{1}{10} \cdot 10 ( 10^3 - 1 ) }{10 -1} + \dots \right )$

$= \dfrac{6}{9} (9 + 99 + 999 + \dots )$

Now:
$= \dfrac{6}{9} ((10 - 1) + (100 - 1) + (1000 - 1) + \dots )$

$= \dfrac{6}{9} ( (10 + 100 + 1000 + \dots ) - (1 + 1 + 1 + \dots ) )$ <-- we have n terms

$= \dfrac{6}{9} \left ( \dfrac{ 10 (10^n - 1)}{10 - 1} - n \right )$

$= \dfrac{60}{81} 10^n - \dfrac{60}{81} - \cdot \dfrac{6}{9} n$

So finally(!)
$6 + 66 + 666 + \dots = \dfrac{6}{81} ( 10 \cdot 10^n - 10 - 9n)$

-Dan

Addendum: While I was working on this, I found a shortcut.
$6 + 66 + 666 + \dots = 6 ( 1 + 11 + 111 + \dots ) = \dfrac{6}{9} \cdot 9 (1 + 11 + 111 + \dots ) = \dfrac{6}{9} (9 + 99 + 999 + \dots )$

and go from there.

 

[Dr Adam]

View attachment 34248Hi Mr Dan,
sorry I got distracted with exam for past few days, but I am still very interested in solving this,
I am stuck in finding the ratio. Is it 10 ? I tried dividing the consecutive terms to find r but it turns out they are not the same?
In your previous post you put 10^k - 1, where is the position of that -1, next to the k? or next to 10^k.
I am dealing with series within series here, but I guess forming the expression is my biggest hurdle..
please help

Hi,

So I know you said you're not terribly familiar with summation notation but now would be a good time and it's not all that bad.

First we need to express the following expression using summation notation,



which as mentioned above would be,

$S_n = 6\left(\sum_{k=1}^{n} \frac{10^k-1}{9} \right)$

Then we can take out the constants and distribute the summation over each term to give,

$[6\left(\sum_{k=1}^{n} \frac{10^k-1}{9} \right)= (6/9) \sum_{k=1}^{n} 10^k-(6/9) \sum_{k=1}^{n} 1=(6/9)\frac{10(10^n-1)}{9}-\frac{6n}{9}=\frac{60(10^n - 1)}{81}-\frac{18n}{27}$

Now, the expression you have has (n + 1) in the exponent so changing it from 'n' to 'n + 1' means multiplying by 10, so to compensate, we can do the same with the '1' by changing it to a '10' then divide the multiplier outside by 10 to give,

$\frac{60(10^n - 1)}{81}-\frac{18n}{27}=\frac{6(10^{(n+1)} - 10)}{81}-\frac{18n}{27}$

Finally we can use 27 in the denominator and bring the 18n into the bracket that is being multiplied by 2 like this,

$\frac{6(10^{(n+1)} - 10)}{81}-\frac{18n}{27}=\frac{2(10^{(n+1)} - 10)}{27}-\frac{18n}{27}=\frac{2(10^{(n+1)} - 10-9n)}{27}$

as required.

 

[nanase]

Ah Thank You Mr. Dan! the little algebra work where you take the -1 out and put them together made it easier for me to see how the next step is formed. Much Thank you and really appreciate the help! I fully understand it now.

 

[Mr. Bland]

I'm a bit confused regarding what the sequence is supposed to be. Is it correct to say that $S_1 = 6$, $S_2 = 66$ and so-on? And the problem is asking for a formula for the sum of all terms from $S_1$ through $S_n$ given some value for $n$?

Assuming I interpret this correctly, my train of thought is as follows...

Each term in the sequence can be given with the following formula:

$S_n = \frac{2}{3}(10^n-1)$​

This isn't actually in the form of a geometric series, as the terms are not separated by a common ratio. 6 to 66 is a ratio of 11, 66 to 666 is a ratio of $10.\overline{09}$, etc.

The running sum of terms is given with the following formula (given that $\Sigma_0=0$):

$\Sigma_n = \Sigma_{n-1} + S_n$​

This is a recurrence relation, and I never did learn how to work those out. Wolfram|Alpha tells me that:

$\Sigma_n = \frac{20}{27}(10^n-1) - \frac{2n}3$​

This solves the problem as I understand it.

I'm interested to know how to resolve the formula properly--how to convert the recursive formula into something that is not recursive. I see this come up a lot in various contexts and it always flies over my head. In a pinch, I don't mind using Wolfram|Alpha for implementing solutions, but I like to know to do things "without a calculator" as it were. Any insights into this would be appreciated.

 

[topsquark]

I'm a bit confused regarding what the sequence is supposed to be. Is it correct to say that $S_1 = 6$, $S_2 = 66$ and so-on? And the problem is asking for a formula for the sum of all terms from $S_1$ through $S_n$ given some value for $n$?

Assuming I interpret this correctly, my train of thought is as follows...

Each term in the sequence can be given with the following formula:

$S_n = \frac{2}{3}(10^n-1)$​

This isn't actually in the form of a geometric series, as the terms are not separated by a common ratio. 6 to 66 is a ratio of 11, 66 to 666 is a ratio of $10.\overline{09}$, etc.

The running sum of terms is given with the following formula (given that $\Sigma_0=0$):

$\Sigma_n = \Sigma_{n-1} + S_n$​

This is a recurrence relation, and I never did learn how to work those out. Wolfram|Alpha tells me that:

$\Sigma_n = \frac{20}{27}(10^n-1) - \frac{2n}3$​

This solves the problem as I understand it.

I'm interested to know how to resolve the formula properly--how to convert the recursive formula into something that is not recursive. I see this come up a lot in various contexts and it always flies over my head. In a pinch, I don't mind using Wolfram|Alpha for implementing solutions, but I like to know to do things "without a calculator" as it were. Any insights into this would be appreciated.

If you know summation notation
$6 + 66 + 666 + \dots = 6(1 + 11 + 111 + \dots )= 6 ( 1 + (1 + 10) + (1 + 10 + 100) + \dots )$

$\displaystyle = 6 \sum_{j=1}^n \left ( \sum_{k = 1}^j 10^{k-1} \right )$

and you can apply the sum rules for a geometric series to that. This is the fastest and easiest way to do the problem if you know how to do it.

-Dan

 

[nanase]

Hi,

So I know you said you're not terribly familiar with summation notation but now would be a good time and it's not all that bad.

First we need to express the following expression using summation notation,

View attachment 34249

which as mentioned above would be,

$S_n = 6\left(\sum_{k=1}^{n} \frac{10^k-1}{9} \right)$

Then we can take out the constants and distribute the summation over each term to give,

$[6\left(\sum_{k=1}^{n} \frac{10^k-1}{9} \right)= (6/9) \sum_{k=1}^{n} 10^k-(6/9) \sum_{k=1}^{n} 1=(6/9)\frac{10(10^n-1)}{9}-\frac{6n}{9}=\frac{60(10^n - 1)}{81}-\frac{18n}{27}$

Now, the expression you have has (n + 1) in the exponent so changing it from 'n' to 'n + 1' means multiplying by 10, so to compensate, we can do the same with the '1' by changing it to a '10' then divide the multiplier outside by 10 to give,

$\frac{60(10^n - 1)}{81}-\frac{18n}{27}=\frac{6(10^{(n+1)} - 10)}{81}-\frac{18n}{27}$

Finally we can use 27 in the denominator and bring the 18n into the bracket that is being multiplied by 2 like this,

$\frac{6(10^{(n+1)} - 10)}{81}-\frac{18n}{27}=\frac{2(10^{(n+1)} - 10)}{27}-\frac{18n}{27}=\frac{2(10^{(n+1)} - 10-9n)}{27}$

as required.

Just noticed this, thank you so much Dr Adam, this brings another light to my understanding. Appreciate it so much!