A Finite Series Workout

(A new question of the week)

A question from the end of August led a student and a Math Doctor to an extra challenge, by way of an apparent typo in the problem. We particularly enjoy working with students who are willing to take on extra work in order to learn more than they need to!

The problem as given

Here is the question:

I have been given the following problem.

Find the sum of

1.2.3 + 2.4.6 + 3.6.12 + ……. to n terms

Here is how I tried,

The rth term is (r(r+1)/2).(2r).(3.2^(r-1)).

I could not do the summation because of the last geometric sequence term. Request you to help me.

One more important thing is that instead of giving the answer, the book has given the formula for the nth term of the sum. It is (3n^(2).(n+1)^(2))/2.

I checked it for the given 3 terms and it works.

Regards,

Rahul.

Rahul is carefully using r as the index of terms in the series, as distinct from n which is the number of terms in the series; others might have used i or k.

Doctor Rick replied:

Hi, Rahul. You write:

Find sum of 1.2.3 + 2.4.6 + 3.6.12 + ……. to n terms

Here is how I tried, The rth term is (r(r+1)/2).(2r).(3.2^(r-1))

I think you may have made an error here by writing the formula for the sum of the n first factors rather than the first factor itself. We can’t be certain about the pattern of the terms from just three terms (or, really, any finite number of terms!), but it seems likely that the first factor of the rth term, 1, 2, 3, …, is just r. Instead you wrote the sum 1 + 2 + 3 + … + r, which is r(r+1)/2.

For the other two factors of each term, we have the sequences 2, 4, 6, … and 1, 2, 4, …, which fit the formulas 2r and 2r – 1. Thus, again, though we can’t be sure this is the intention, it’s a reasonable assumption given that we are forced to make an assumption!

So a corrected formula for the rth term of the series, as it appears to be, is $$a_r = (r)(2r)(3\cdot2^{r-1}) = 3r^2 2^{r}$$

One more important thing is that instead of giving the answer, the book has given the formula for the nth term of the sum. It is (3n^(2).(n+1)^(2))/2. I checked it for the given 3 terms and it works.

The first point to notice here is that the problem is looking for the formula for the sum of the first n terms of the series. This is often denoted as Sn, the nth term of the sequence of partial sums of the series. If the book’s answer calls it “the nth term of the sum”, it must mean the nth term of the sequence of partial sums. It surely doesn’t mean the nth term of the series.

So the answer the book gives is presumably just what it asks for, though the wording taken literally would mean the nth term \(a_n\) of the series (summation), which would be our formula above. The book is saying that $$S_n = \frac{3n^2(n+1)^2}{2}$$ But is that right?

My next observation is that when I check this formula “for the given 3 terms”, I do not find that it works. We have

a1 = 1*2*3  =   6     …   S1 =  6

a2 = 2*4*6  =  48    …   S2 =  6 +  48 =  54

a3 = 3*6*12 = 216  …   S3 = 54 + 216 = 270

while from the book’s formula I get

Sn = (3n2)(n+1)2/2

S1 = (3*12)(1+1)2/2 =   6

S2 = (3*22)(2+1)2/2 =  54

S3 = (3*32)(3+1)2/2 = 216

It checks out for the first two partial sums, but not for the third. Did you get different results?

So either the terms shown in the problem are wrong, or the formula given for the sum is wrong! But look at the book’s formula for the sum: Does it look right for a sum of terms as given by our formula?

A proposed correction

Now, the fact that your formula for the rth term of the series has a geometric component (quite reasonably, since the sequence 1, 2, 4, … is the start of a geometric sequence), while the book’s formula does not look geometric at all, suggests that either we need to assume some other pattern that also starts 1, 2, 4, …, or that the series given in the problem has a typographical error in it! I think the latter is likely, since as I showed, the book’s answer does not work for the series given.

As it turns out, when I first saw the problem I misread it as

1·2·3 + 2·4·6 + 3·6·9 + …

I suppose I did that because I was subconsciously making it a much easier problem — and in fact, the sum of this series is given by the book’s answer. Thus my theory is that there is a typo in the problem, and that it was intended to have 9 instead of 12.

What do you think of this?

Changing one number in the problem makes the book’s answer correct, so that seems a likely fix.

Rahul answered,

Dear Sir,

There seems to be typing error in the book about the answer or the question itself.

The problem is as follows,

Find the sum

1*2*3 + 2*4*6 + 3*6*12 + ……. up to n terms.

The answer given in the book also is not right. As you pointed out my mistake also, it is neither the formula for the sum up to n terms nor the formula for nth term.

Recognizing that there is an error, Rahul now showed in detail how to obtain the formula for the rth term in the series as given, correcting his original attempt (but not, yet, the problem itself):

In fact my work should have been as follows,

The first factor in the first term is 1, and the same in the second term is 2, and the same in the third term is 3, so I thought it is an arithmetic sequence with a=1, d=1, so I used the formula for its nth term as the general formula for the first factor of the nth (or the rth term before summing all the n terms).

It was 1+(r-1)*1 = r

The second factor in the first term is 2, and the same in the second term is 4, and the same in the third term is 6, so I thought it is an arithmetic sequence with a=2, d=2, so I used the formula for its nth term as the general formula for the first factor of the nth (or the rth term before summing all the n terms).

It was 2+(r-1)*2 = 2r.

The third factors were difficult to understand at first sight as it were not clearly arithmetic progression.

So I used the formula for Geometric progression and found the 3rd factor of the rth term as aq^(r-1) where q = 6/3 = 12/6=2. So it was 3*2^(r-1).

The product of these three factors is the formula I showed above for the rth term, \(a_r = (r)(2r)(3\cdot2^{r-1}) = 3r^2 2^{r}\). But this is evidently not the series the author intended.

Solving the corrected problem

So let’s assume the problem is really this:

Find the sum \(1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+\dots\) up to n terms.

The rth term is then \((r)(2r)(3r) = 6r^3\), which is much simpler.

Rahul continued:

So finally,

There is typing error in the problem, if the answer works for this correct problem instead 1·2·3 + 2·4·6 + 3·6·9 + …

And yes the answer is for this last sum itself.

That is Sum (r goes from 1 to n) of 6*r^(3) = (3*n^(2)*(n+1)^(2))/2.

Thank you so much for the help and revealing what should have been the correct problem itself.

Rahul hasn’t shown the work for obtaining this formula; I’m not sure what method he is expected to use; but the sum of cubes is a standard formula one can memorize or look up, which may be all that he did: $$\sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}$$ Memorably, this is the square of the arithmetic series: $$\sum_{r=1}^n r^3 = \left(\sum_{r=1}^n r\right)^2$$

For explanations of ways to obtain this and similar formulas (mostly using sums of other powers as examples), see:

Summing Integers to the Fourth Power

Finding Sum Formula using Sequences of Differences

Induction Problem

Sum of Consecutive Squares

Sigma Notation

Can we solve the mis-stated problem?

Doctor Rick said,

Good, I believe you are satisfied with the idea that there was an error in the problem or the solution. It can be fun to take on the challenge of figuring out what an erroneous problem was likely intended to be, and this helps convince us that it isn’t our fault that we don’t get the “correct” answer!

If you would still like to solve the problem as stated, even though it doesn’t seem to be the intended problem and is significantly more difficult, I believe the method(s) given in this item from the Ask Dr. Math Archive will be of help:

Finding the Sum of Arithmetico-Geometric Series

Here is the question and the initial answer from the page:

Finding the Sum of Arithmetico-Geometric Series

Find the sum of the infinite series

  1/7 + 4/(7^2) + 9/(7^3) + 16/(7^4) +...

I would also like to know if there is a general rule to find the sum of (n^2/p^n) for n = 1 to infinity.

The series looks like a convergent sequence.  I wonder if there is a  general rule to break a convergent sequence like above into one or more convergent sequences?

Though this is an infinite series, not a finite sum like our problem, then terms are quite similar to ours.

Doctor Vogler’s answer:

Hi Sudheer,

Thanks for writing to Dr. Math.  Have you ever summed an "arithmetico-geometric series"?  That is, one of the form

       k
  S = sum a * n * r^n  ?
      n=1

Some people do this by taking derivatives, but I won't assume that you know calculus, and I'll do it the algebraic way:  Subtract r*S.

             k                 k
  S - r*S = sum a * n * r^n - sum a * n * r^(n+1)
            n=1               n=1

             k                k+1
  S - r*S = sum a * n * r^n - sum a * (n-1) * r^n
            n=1               n=2

                            k+1
  S - r*S = a * 1 * r^1 + [ sum a * r^n ] - a * k * r^(k+1)
                            n=2

and then the middle series is a geometric series, and I assume that you can sum a series like that.  This gives you a formula for S - r*S, and then you just divide both sides by 1-r to get S.

Well, you have a series of the form

       k
  S = sum a * n^2 * r^n
      n=1

In fact, you can sum it in exactly the same way:  Subtract r*S, regroup the sums, divide by 1-r, and you will get a formula for S which has an arithmetico-geometric series on the right, the type that I already solved.  Use the formula for that kind of series to get the formula for your kind of series.

Note that this first defines an arithmetico-geometric series, whose terms are the product of an arithmetic sequence and a geometric sequence, and shows the technique for solving it; then he states that the problem posed is not that, but can be solved similarly.

Also, don’t let it confuse you that he uses r and n differently than in our problem; here n is the index in the series (our r), k is the number of terms (our n), and r is the common ratio (our 2).

Taking the challenge of the typo

Rahul answered, not lingering on the apparently correct problem, but leaping to the challenge of the problem as written:

Thanks for the reply.

I am taking a look at the Arithmetico-Geometric Series and definitely try to solve the as stated problem.

The book also mentions about the Arithmetico-Geometric series.

I am not able to show that it is arithmetico-geometric series because the first two factors of each term do not multiply to give an arithmetic sequence.

I mean it comes out to me as 2*3 + 8*6 + 18*12 + …. to n terms.

And here the factors 2, 8, 18 do not give me an arithmetic progression so as to use the formula for a general arithmetico-geometric series like, (a+(n-1)*d)*(g*r^(n-1)).

I have the first factor 3*2^(n-1) but not the first one.

Request you to help.

Doctor Rick corrected the misunderstanding,

The title of the link I provided is somewhat misleading, because it is not only about arithmetico-geometric series. The initial question is about a series more like ours — the term-by-term product of a quadratic series and a geometric series. Doctor Vogler first reviews summation methods for an arithmetico-geometric series, and then shows how to change our kind of series into that:

Well, you have a series of the form

     k
S = sum a * n^2 * r^n
    n=1

In fact, you can sum it in exactly the same way:  Subtract r*S, regroup the sums, divide by 1-r, and you will get a formula for S which has an arithmetico-geometric series on the right, the type that I already solved.  Use the formula for that kind of series to get the formula for your kind of series.

Can you follow this guidance? I haven’t tried it myself yet.

A typo of our own

Rahul responded, having first tried to replicate Doctor Vogler’s work before moving to his own:

Dear sir,

I tried to repeat Dr. Vogler’s work in that reply but I am getting the different answer after simplification as follows

a1r + ( sum from n = 2 to k (instead of k+1) of ar^(n) ) – akr^(k+1).

Can you please help?

Request you to help only for this step. Then again I will move forward

Doctor Rick answered,

I believe you’re referring to the derivation of the last line in the following part of Dr. Vogler’s first response:

             k                 k
  S - r*S = sum a * n * r^n - sum a * n * r^(n+1)
            n=1               n=1

             k                k+1
  S - r*S = sum a * n * r^n - sum a * (n-1) * r^n
            n=1               n=2

                            k+1
  S - r*S = a * 1 * r^1 + [ sum a * r^n ] - a * k * r^(k+1)
                            n=2

I didn’t think you would need to work on this part, since you indicated that your book “mentions” arithmetico-geometric series. I guess it doesn’t fully teach them, or else you haven’t studied that part. (I have never studied this topic myself.)

In going from the second line to the third, we want to combine the two sums; but in order to do that, we must first pull out terms that they do not have in common — the n=1 term in the first, namely a*1*r^1 = ar, and the n=k+1 term in the second, namely a*r^(k+1). This leaves summations from n = 2 to k, and subtracting term by term, we obtain sum[n=2 to k] (a*n*r^n – a(n-1)r^n), or sum[n=2 to k] a*r^n.

So I agree with you, there is an error in Dr. Vogler’s work. Unfortunately, we can no longer request corrections in the Ask Dr. Math Archive.

To correct Doctor Vogler’s page, we would need to change the upper limit of the summation in the third line from k + 1 to k. But that’s a minor typo.

Solving the quadratic-geometric series

In any case, I would not use what Dr. Vogler wrote at all — I’m not interested in obtaining a general formula. (If I worked regularly with such series, I might want that!) All I’d do is to apply the method to the specific series of interest.

What arithmetico-geometric series will we want to sum? As Dr. Vogler suggests, we can apply the same idea to our quadratic-geometric series. Again, there is no need (as far as I am concerned) for generality. We want to sum the series

Σ[k=1 to n] (6k2 2k-1) = 3 Σ[k=1 to n] k2 2k

We will just look at the last sum, ignoring the multiplication by 3. Call it S:

S = Σ[k=1 to n] k2 2k = 12 21 + Σ[k=2 to n] k2 2k = 2 + Σ[k=2 to n] k2 2k

2S = Σ[k=1 to n] k2 2·2k = Σ[k=1 to n] k2 2k+1 = Σ[k=2 to n+1] (k–1)2 2k

     = Σ[k=2 to n] (k–1)2 2k + n2 2n+1

2S – S = n2 2n+1 – 2 + Σ[k=2 to n]((k–1)2 – k2) 2k

S = n2 2n+1 – 2 + Σ[k=2 to n](1 – 2k) 2k

That’s as far as I’ll take it (and I don’t guarantee I made no mistakes myself). The sum in the last expression for S is the difference of a geometric series and twice the arithmetico-geometric series Dr. Vogler discussed.

He has modified the procedure a bit, for example subtracting S from 2S rather than vice verse for convenience. I made a couple small corrections here, so if there is an error, it may be due to either of us.

Rahul answered,

Dear Sir,

Thanks for making me work hard.

After a lot of work I finally found the answer, here it is,

The sum S = 3*(2^(n+1)-4-2*(n*2^(n+1)-2(2^(n)-1))+4-2+n^(2)*2^(n+1)).

This time I am sure that I have checked it also correctly.

It was really a difficult fun.

I am very happy that I could learn lot of stuff unknown to me earlier.

Thank you very much for the guidance.

I like that phrase, “a difficult fun”. I tell my students that my definition of “fun” is “challenging”, so they’re ready when I end a session with, “Wasn’t that fun?” I’m not going to try to fill in all the gaps; that would take away some of the fun for you, the reader.

Let’s check that \(S_3 = 270\), as we found from the three given terms:

$$S_3 = 3\left(2^{3+1}-4-2\left(3\cdot2^{3+1}-2\left(2^{3}-1\right)\right)+4-2+3^22^{3+1}\right) \\= 3\left(16-4-2\left(3\cdot16-2\left(8-1\right)\right)+4-2+9\cdot16\right) \\= 3\left(16-4-2\left(3\cdot16-14\right)+4-2+9\cdot16\right) \\= 3\left(16-4-68+4-2+144\right) = 3(90) = 270$$

Doctor Rick replied,

I’m glad you enjoyed the challenge! You have learned more than I did, because you did more of the work.

I checked your formula by putting it into a spreadsheet, and it agrees with the directly calculated sum as far as Excel can calculate exactly, which is n = 22. This strongly suggests that you solved the problem correctly. We can simplify it:

S[n] = 6(n^2 – 2n + 3)2^n – 18

Simplification isn’t essential (Rahul just wanted to get to an answer he could check), but it does make it look more like a final answer:

$$1\cdot 2\cdot 3+2\cdot 4\cdot 6+3\cdot 6\cdot 12+\dots \text{ (n terms) }= \sum_{k=1}^n 3k^2 2^{k} = 6\left(n^2 – 2n + 3\right)2^n – 18$$

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