Question

What is the sum of `1^5 + 2^5 + 3^5 + … + 20^5?`

Answer 1

`12333300`

Explanation:

Write `s_n` for the sum to `n` terms.

Note that the general term `a_n = n^5` is quintic in `n`. Hence the sum will be of degree `6` and therefore determined by `7` coefficients. It therefore suffices to calculate `s_1,... s_7` and apply a method of differences to derive a formula for `s_n`:

`s_1 = 1^5 = 1`

`s_2 = s_1 + 2^5 = 1+32 = 33`

`s_3 = s_2 + 3^5 = 33+243 = 276`

`s_4 = s_3 + 4^5 = 276+1024 = 1300`

`s_5 = s_4 + 5^5 = 1300+3125 = 4425`

`s_6 = s_5 + 6^5 = 4425+7776 = 12201`

`s_7 = s_6 + 7^5 = 12201+16807 = 29008`

Writing these values as a sequence, we have:

`color(blue)(1), 33, 276, 1300, 4425, 12201, 29008`

Then write down the sequence of differences between consecutive terms:

`color(blue)(32), 243, 1024, 3125, 7776, 16807`

Write down the sequence of differences of those differences:

`color(blue)(211), 781, 2101, 4651, 9031`

Write down the sequence of differences of those differences:

`color(blue)(570), 1320, 2550, 4380`

Write down the sequence of differences of those differences:

`color(blue)(750), 1230, 1830`

Write down the sequence of differences of those differences:

`color(blue)(480), 600`

Write down the sequence of differences of those differences:

`color(blue)(120)`

Then we can use the first term of each sequence we found as coefficients for a formula for `s_n`:

`s_n = color(blue)(1)/(0!)+color(blue)(32)/(1!)(n-1)+color(blue)(211)/(2!)(n-1)(n-2)+color(blue)(570)/(3!)(n-1)(n-2)(n-3)+color(blue)(750)/(4!)(n-1)(n-2)(n-3)(n-4)+color(blue)(480)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)+color(blue)(120)/(6!)(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)`

`color(white)(s_n) = 1+32n-32+211/2n^2-633/2n+211+95n^3-570n^2+1045n-570+125/4n^4-625/2n^3+4375/4n^2-3125/2n+750+4n^5-60n^4+340n^3-900n^2+1096n-480+1/6n^6-7/2n^5+175/6n^4-245/2n^3+812/3n^2-294n+120`

`color(white)(s_n) = 1/12n^2(2n^4+6n^3+5n^2-1)`

Hence:

`s_20 = 1/12(color(blue)(20))^2(2(color(blue)(20))^4+6(color(blue)(20))^3+5(color(blue)(20))^2-1)`

`color(white)(s_20) = 400/12(2(160000)+6(8000)+5(400)-1)`

`color(white)(s_20) = 100/3(320000+48000+2000-1)`

`color(white)(s_20) = 100/3(369999)`

`color(white)(s_20) = 12333300`