Question

Find the sum of series whose nth term is given by `n( n + 1) ( n + 4)`?

Answer 1

`s_n = 1/12n(n+1)(n+2)(3n+17)`

Explanation:

Given:

`a_n = n(n+1)(n+4)`

Note that this is a cubic formula, so the formula for the sum to `N` terms will be a quartic polynomial in `N`.

We can find it using a method of differences.

One advantage of this method is that you do not have to memorise formulas for `sum n`, `sum n^2`, `sum n^3`, etc.

The first few terms of the given sequence are:

`a_1 = 1 * 2 * 5 = 10`

`a_2 = 2 * 3 * 6 = 36`

`a_3 = 3 * 4 * 7 = 84`

`a_4 = 4 * 5 * 8 = 160`

`a_5 = 5 * 6 * 9 = 270`

`a_6 = 6 * 7 * 10 = 420`

`a_7 = 7 * 8 * 11 = 616`

(Note that I have calculated a couple more terms here than is really necessary, but it makes the analysis below more "compelling").

Writing down the sequence of the first few sums (`s_n`), we have:

`color(blue)(10), 46, 130, 290, 560, 980, 1596`

Then the sequence of differences between consecutive terms is:

`color(blue)(36), 84, 160, 270, 420, 616`

The sequence of differences of those differences is:

`color(blue)(48), 76, 110, 150, 196`

The sequence of differences of those differences is:

`color(blue)(28), 34, 40, 46`

The sequence of differences of those differences is:

`color(blue)(6), 6, 6`

Notice that we arrived at a constant sequence after `4` steps of "taking the difference". This is what we expect of a sequence of terms described by a quartic formula.

Having arrived at a constant sequence, we can use the initial terms of each of the sequences as coefficients to write a polynomial matching the sample points:

`s_n = color(blue)(10)/(0!)+color(blue)(36)/(1!)(n-1)+color(blue)(48)/(2!)(n-1)(n-2)+color(blue)(28)/(3!)(n-1)(n-2)(n-3)+color(blue)(6)/(4!)(n-1)(n-2)(n-3)(n-4)`

`color(white)(s_n) = 10+36n-36+24n^2-72n+48+14/3n^3-28n^2+154/3n-28+1/4n^4-5/2n^3 +35/4n^2-25/2n+6`

`color(white)(s_n) = 1/12n(3n^3+26n^2+57n+34)`

`color(white)(s_n) = 1/12n(n+1)(3n^2+23n+34)`

`color(white)(s_n) = 1/12n(n+1)(n+2)(3n+17)`