Use summation identities to rewrite the given expression without using summation notation.

Question

n

∑ i(i-9)

i=1 n³

Answer 1

Hi Sarah S.,

I think what this question is getting at is the polynomial expression of the particular series. This resembles the indefinite integral of the corresponding algebraic expression (here, i^2 - 9i ; the n^3 in the denominator is a one-time-applied scaling factor on the result). However, it is a little different in that it is "grainy", like integrating by stairsteps which you recall from beginning calculus exercises. That is in the nature of a polynomial expression, which you are evaluating ONLY at the integer values.

So how do you go about finding that polynomial expression? The easiest way is by writing the particular series as the first column in a table (1=1, 2, 3....), then with Sigma (running total) as the next column. Also do this for the second column = i^2, and the third column = Sigma (running total of those). Then start reducing the Sigma columns by successive differences. A convenient name for this is "decomposing the series by successive differences". I'll show you what this looks like as a horizontal table, because that's all I can conveniently type here in Wyzant!

i column: 1 2 3 4 5 6 7 .....

Sigma(i) 1 3 6 10 15 21 28 .....

1st diff 1 2 3 4 5 6 7 .... (note that you got the original i column back!)

2nd diff 1 1 1 1 1 1 1

Note that the 2nd difference column is all a constant value. By comparison, check that behavior (2nd diff = constant, value = 1) against the behavior of terms of a polynomial, similarly deomposed. Look first at what an x^1 term gives:

i column 1 2 3 4 5 6 7 ....

1st diff 1 1 1 1 1 1 1 ....

Note that an x^1 term yields a constant value of 1 for the **first** difference, NOT for the 2nd difference!

Now try it for an x^2 term:

i column 1 4 9 16 25 36 49 ....

1st diff 1 3 5 7 9 11 13 ...

2nd diff 2 2 2 2 2 2 ...

Note that an x^2 term, so decomposed, yields a constant for the 2nd difference, BUT that constant value is 2, not 1 as above for SIGMA(i) . Therefore you know instantly that the behavior of an (x^2)/2 term will scale down to the same decomposition behavior as did the SIGMA(i) series, and therefore the expression for Sigma(i) series AS A POLYNOMIAL has a largest-exponent term of (x^2)/2.

But you aren't quite done yet -- that was just the largest exponent term. There may still be terms of lesser exponents "hidden" in the actual series for Sigma(i), so what you have to do next is to subtract the (x^2)/2 term from the corresponding Sigma(i) term, term by term, to get the "remainder" series:

Sigma(i) series 1 3 6 10 15 21 28 ....

(x^2)/2 series 0.5 2 4.5 8 12.5 18 24.5 ....

remainder 0.5 1 1.5 2 2.5 3 3.5 ...

1st diff 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...

Viola! 1st difference "does" it; compare that with the constant "1" for the x^1 series above, to see that you also have an (x/2) term in the desired polynomial you are building, which is now (x^2)/2 + (x/2).

If you now generate the terms for the (x^2)/2 +(x/2) series, you will see that they exactly match the desired series for Sigma(i) above. So you have done it (for part of the desired expression you are solving, recall that was i^2 - 9i in all . In particular, you have the part for the -9i (yes, you still have to multiply by -9/(n^3), but that is the easy part)).

But .... you still have the harder part ahead ... doing the same thing for the Sigma (i^2) series. Here, you will need to go out to **3rd** differences on the decomposition, and you will end up with terms in x^3, x^2, and x. Once you get those three terms (I think that's it, though there might be a constant added also), you add them to the corresponding terms for the Sigma(i) etc. part, and crank out your final expression form.

Sarah, I realize this may seem like a huge rigmarole to you right now, but write the things above as vertical column tabular format instead, and the idea of successive differences will pop out at you.

One other thing: you will realize (perhaps) that the first term of the polynomial expression for the summation series of each of the powers-of-x terms looks an awful lot like the indefinite integral of that function of x. And so it is! So why go to so much trouble to show this in a calculus course? Sometimes being able to manipulate a complex expression as such polynomial series enables you to argue things such as behaviors of the expressions as the variable goes to a limit value. You may have been exposed to this in l'Hopital's Rule, but polynomials are an alternate way of viewing such things that can make them "make sense".

-- Cheers, --Mr. d.

P.S. If you want to go to the trouble of writing the polynomial series for Sigma (x^n) where n is larger integers, you will see some intriguing patterns emerging from the successive series -- you can eventually start writing the expressions out from the series just "before" -- except for the pesky last term. It's a real "doozy" (that used to be spelled, "Deusie", do you know why? Look up: Deusenberg).