What is #1^2+(1^2+2^2)+(1^2+2^2+3^2)+...+(1^2+2^2+...+22^2)# ?
3 Answers
By observation we get the
Putting
Explanation:
Here's an alternative method based on differences.
Note that each "term" is a sum of quadratic terms, so will be given by a cubic formula.
Then the sum of all of the "terms" will have a quartic formula.
A quartic polynomial is determined by
#1^2 = 1#
#1^2+(1^2+2^2) = 1+5 = 6#
#1^2+(1^2+2^2)+(1^2+2^2+3^2) = 1+5+14 = 20#
#1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2) = 1+5+14+30 = 50#
#1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+(1^2+2^2+3^2+4^2+5^2) = 1+5+14+30+55 = 105#
Write out this sequence:
#color(blue)(1), 6, 20, 50, 105#
Write out the sequence of differences between consecutive terms:
#color(blue)(5), 14, 30, 55#
Write out the sequence of differences of those differences:
#color(blue)(9), 16, 25#
Write out the sequence of differences of those differences:
#color(blue)(7), 9#
Write out the sequence of differences of those differences:
#color(blue)(2)#
Hence the formula for the sum to
#s_n = color(blue)(1)/(0!)+color(blue)(5)/(1!)(n-1)+color(blue)(9)/(2!)(n-1)(n-2)+color(blue)(7)/(3!)(n-1)(n-2)(n-3)+color(blue)(2)/(4!)(n-1)(n-2)(n-3)(n-4)#
#color(white)(s_n) = 1+5n-5+9/2n^2-27/2n+9+7/6 n^3-7n^2+77/6n-7+1/12n^4-5/6n^3+35/12n^2-25/6n+2#
#color(white)(s_n) = 1/12n(n^3+4n^2+5n+2)#
Then:
#s_22 = 1/12(color(blue)(22))((color(blue)(22))^3+4(color(blue)(22))^2+5(color(blue)(22))+2) = 23276#
See below.
Explanation:
First we will determine a general formula for the sum
Giving values we have
This set of equations is easily solved giving