Question

What is the sum of the sequence `2, 5, 10, 17,...` up to `n` ?

Answer 1

`a_n = n^2+1`

`s_n = 1/6n(2n^2+3n+7)`

`s_(a_n) = ...`

Explanation:

We are not told what kind of series this is, so we can only attempt to infer from the terms we are given. Note that no finite sequence determines the terms that follow it, unless you are given additional information.

The given example is neither an arithmetic sequence (with common difference) nor a geometric sequence (with common ratio).

I can see that the differences increase linearly, so the terms fit a quadratic formula and their sums will fit a cubic formula.

Writing out the original sequence, we have:

`color(blue)(2), 5, 10, 17`

Write out the sequence of differences between consecutive terms:

`color(purple)(3), 5, 7`

Write out the sequence of differences of that sequence:

`color(violet)(2), 2`

Having arrived at a constant sequence, we can write a formula for the `n`th term of the original sequence, using the first term of each of the above sequences as coefficients:

`a_n = color(blue)(2)/(0!)+color(purple)(3)/(1!)(n-1)+color(violet)(2)/(2!)(n-1)(n-2)`

`color(white)(a_n) = 2+color(red)(cancel(color(black)(3n)))-3+n^2-color(red)(cancel(color(black)(3n)))+2`

`color(white)(a_n) = n^2+1`

We can similarly derive a formula for the sum to `n` terms, `s_n`, as follows:

Write down the first few terms:

`color(blue)(2), 7, 17, 34`

Write down the sequence of differences:

`color(purple)(5), 10, 17`

Write down the sequence of differences:

`color(violet)(5), 7`

Write down the sequence of differences:

`color(red)(2)`

Having arrived at a constant sequence (albeit of just one term), we can write down our formula:

`s_n = color(blue)(2)/(0!)+color(purple)(5)/(1!)(n-1)+color(violet)(5)/(2!)(n-1)(n-2)+color(red)(2)/(3!)(n-1)(n-2)(n-3)`

`color(white)(s_n) = 2+5n-5+5/2n^2-15/2n+5+1/3n^3-2n^2+11/3n-2`

`color(white)(s_n) = 1/3n^3+1/2n^2+7/6n`

`color(white)(s_n) = 1/6n(2n^2+3n+7)`

Note that the question asks for the sum to the term with value `n` rather than the sum of `n` terms. We can derive this from our formula for `a_n` by taking its inverse:

`s_(a_n) = 1/6n(2n^2+3n+7)`

So the sum to value to `n` is:

`1/6k(2k^2+3k+7)`

where `k = sqrt(n-1)`