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

1 Answer
Aug 14, 2018

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