Question

Derive the Formula `sum_(k=1)^nk^4=1/30(6n^5+15n^4+10n^3-n)`?

Not even sure where to start. Any guidance would be greatly appreciated.

Thanks!

Answer 1

Consider first the following sum:

`sum_(k=1)^n (k+1)^5-k^5`

This is a sum of differences, and if we expand terms we can quickly establish the formula for this sum, as almost all the terms cancel:

`sum_(k=1)^n (k+1)^5-k^5 = (color(red)(2^5)-1^5) + (color(blue)(3^5)-color(red)(2^5))+ (color(green)(4^5)-color(blue)(3^5)) + (5^5-color(green)(4^5))+... + (n+1)^5-n^5`

After cancelling we are left with:

`\ \ \ \ \ sum_(k=1)^n (k+1)^5-k^5 = -1^5 + (n+1)^5`
`:. sum_(k=1)^n (k+1)^5-k^5 = (n+1)^5 -1 \ \ ... (star)`

Now let us expand the terms on either side using the Binomial Theorem:

On the LHS we have:

`(k+1)^5-k^5 = (1+5k+10k^2+10k^3 +5k^4+k^5) -k^5`
`" " = 1+5k+10k^2+10k^3 +5k^4`

Similarly, on the RHS we have:

`(n+1)^5 -1 = (1+5n+10n^2+10n^3 +5n^4+n^5) -1`
`" " = 5n+10n^2+10n^3 +5n^4+n^5`

Combining these results into `(star)` we can write:

`sum_(k=1)^n (1+5k+10k^2+10k^3 +5k^4) = 5n+10n^2+10n^3 +5n^4+n^5`

And so:

`sum_(k=1)^n 1 + 5sum_(k=1)^nk + 10sum_(k=1)^nk^2 + 10sum_(k=1)^nk^3 + 5sum_(k=1)^nk^4 = 5n + 10n^2 + 10n^3 + 5n^4 + n^5`

We can now use the known standard summation formula:

`sum_(k=1)^n \ k = 1/2n(n+1)`
`sum_(k=1)^n k^2 = 1/6n(n+1)(2n+1)`
`sum_(k=1)^n k^3 = 1/4n^2(n+1)^2`

And then we get:

`n + 5 1/2 n(n+1) + 10 1/6 n(n+1)(2n+1) + 10 1/4 n^2(n+1)^2 + 5sum_(k=1)^nk^4 = 5n + 10n^2 + 10n^3 + 5n^4 + n^5`

Multiplying by `6` we get:

`6n + 15n(n+1) + 10n(n+1)(2n+1) + 15n^2(n+1)^2 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5`

Now multiply out the various brackets:

`6n + 15(n^2+n) + 10n(2n^2+3n+1) + 15n^2(n^2+2n+1) + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5`

`:. 6n + 15n^2+15n + 20n^3+30n^2+10n + 15n^4+30n^3+15n^2 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5`

Collect terms on LHS and RHS:

`:. 31n + 60n^2 + 50n^3 + 15n^4 + 30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5`

Now isolate the sum by collection further terms:

`30sum_(k=1)^nk^4 = 30n + 60n^2 + 60n^3 + 30n^4 + 6n^5 - 31n - 60n^2 - 50n^3 - 15n^4`

`:. 30sum_(k=1)^nk^4 = -n + 10n^3 + 15n^4 + 6n^5`

`:. sum_(k=1)^nk^4 = 1/30(6n^5+ 15n^4+10n^3 -n)` QED

Answer 2

Examine sequences of differences to find the formula...

Explanation:

Since the sum is of polynomial terms of degree `4`, the formula for the sum will be a polynomial of degree `5`.

If the terms of a sequence are given by a polynomial of degree `n`, then forming the sequence of differences, then of differences of differences `n` times will result in a constant sequence.

So the sequence of sums, being a polynomial of degree `5` will reduce to a constant sequence when we take differences `5` times.

To prepare, let us write down the first few powers of `4` (six terms will be just enough):

`1, 16, 81, 256, 625, 1296`

Now write the sequence of the first `6` sums:

`color(blue)(1), 17, 98, 354, 979, 2275`

Next write the sequence of differences between pairs of successive terms:

`color(blue)(16), 81, 256, 625, 1296`

Write down the sequence of differences of those differences:

`color(blue)(65), 175, 369, 671`

Write down the sequence of differences of those differences:

`color(blue)(110), 194, 302`

Write down the sequence of differences of those differences:

`color(blue)(84), 108`

Write down the sequence of differences of those differences:

`color(blue)(24)`

We can now take the initial terms of each of these sequences as coefficients for a formula for the `n`th term of the sequence of sums:

`s_n = color(blue)(1)/(0!)+color(blue)(16)/(1!)(n-1)+color(blue)(65)/(2!)(n-1)(n-2)+color(blue)(110)/(3!)(n-1)(n-2)(n-3)+color(blue)(84)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(24)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)`

`color(white)(s_n) = 1+16n-16+65/2n^2-195/2n+65+55/3n^3-110n^2+605/3n-110+7/2n^4-35n^3+245/2n^2-175n+84+1/5n^5-3n^4+17n^3-45n^2+274/5n-24`

`color(white)(s_n) = 1/5n^5 + 1/2n^4 + 1/3n^3 - 1/30n`

`color(white)(s_n) = 1/30(6n^5 +15n^4+10n^3-n)`

Answer 3

The number-crunching answer.

Explanation:

Making the hypothesis

`sum_(k=1)^n k^4 = c_5 n^5+c_4 n^4+c_3 n^3+c_2 n^2+c_1 n + c_0`

we have

`{(k=1->c_0 + c_1 + c_2 + c_3 + c_4 + c_5 = 1),(k=2->c_0 + 2 c_1 + 4 c_2 + 8 c_3 + 16 c_4 + 32 c_5=17),(k=3->c_0 + 3 c_1 + 9 c_2 + 27 c_3 + 81 c_4 + 243 c_5=98),(k=4->c_0 + 4 c_1 + 16 c_2 + 64 c_3 + 256 c_4 + 1024 c_5=354),(k=5->c_0 + 5 c_1 + 25 c_2 + 125 c_3 + 625 c_4 + 3125 c_5=979),(k=6->c_0 + 6 c_1 + 36 c_2 + 216 c_3 + 1296 c_4 + 7776 c_5=2275):}`

now solving for `c_0,c_1,c_2,c_3,c_4,c_5` we obtain

`c_0= 0, c_1=-1/30, c_2 = 0, c_3 = 1/3, c_4 = 1/2, c_5 = 1/5`

or

`sum_(k=1)^n k^4=-n/30 + n^3/3 + n^4/2 + n^5/5 = 1/30 n (1 + n) (1 + 2 n) (3 n (1 + n)-1)`