Question

What is `1^2+(1^2+2^2)+(1^2+2^2+3^2)+...+(1^2+2^2+...+22^2)` ?

Answer 1

By observation we get the `nth` term of the given series as

`t_n=1^2+2^2+3^2+......+n^2`

`=>t_n=1/6[n(n+1)(2n+1)]`

`=>t_n=1/6[(n^2+n)(2n+1)]`

`=>t_n=1/6[2n^3+3n^2+n]`

`=>S_r=sum_(n=1)^(n=r)t_n=1/6sum_(n=1)^(n=r)[2n^3+3n^2+n]`

`=1/6[2*((r(r+1))/2)^2+3/6(r(r+1)(2r+1))+(r(r+1))/2]`

`=1/12(r(r+1))[r(r+1)+(2r+1)+1]`

Putting `r=22` we get

`S_22=(22*23)/12(22*23+45+1)=23276`

Answer 2

`23276`

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 `5` points, so let us first find the first five terms of the given sequence:

`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 `n` terms is:

`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`

Answer 3

See below.

Explanation:

First we will determine a general formula for the sum

`S_n = sum_(k=1)^n(n-k+1)k^2` or

`S_n = a n^4+bn^3+cn^2+dn+e`

Giving values we have

`{(S_(-2) =16 a - 8 b + 4 c - 2 d + e=0 ),(S_(-1) =a - b + c - d + e=0 ),(S_0=e=0),(S_1 = a + b + c + d + e= 1),(S_2 = 16 a + 8 b + 4 c + 2 d + e = 6):}`

This set of equations is easily solved giving

`a = 1/12, b = 1/3, c = 5/12, d=1/6, e = 0` or

`S_n = sum_(k=1)^n(n-k+1)k^2 = 1/12 n (n+1)^2 (n+2)` and finally

`S_22 = 23276`