Find the value of this expression:
1*22+2*32+3*42+..........+2023*20242
-------------------------------------
12*2+22*3+32*4+..........+20232*2024
Generalize to n terms, then the specific quotient asked for by the problem is the n=2023 value of the generalization.
The numerator is then 1*2^2 + 2*3^2 + ... + n*(n+1)^2
The denominator is then 1^2*2 + 2^2*3 + ... + n^2*(n+1)
Rewrite each term in the numerator as k*(k+1)^2 into k^3 + 2*k^2 + k. Then the numerator can be rearranged into (sum of first n cubes) + 2*(sum of first n squares) + (sum of first n natural numbers)
Rewrite each term in the denominator k^2*(k+1) into k^3 + k^2. Then the numerator can be rearranged into (sum of first n cubes) + (sum of first n squares).
Now we know the formula for the first n cubes is n^2*(n+1)^2/4, the formula for the first n squares is n*(n+1)*(2n+1)/6, and the formula for the first n natural numbers is n*(n+1)/2.
Then the numerator simplifies to n^2*(n+1)^2/4 + n*(n+1)*(2n+1)/3 + n*(n+1)/2 = n*(n+1)*(n+2)*(3n+5)/12
And the denominator simplifies to n^2*(n+1)^2/4 + n*(n+1)*(2n+1)/6 = n*(n+1)*(n+2)*(3n+1)/12
Then the quotient has a lot of stuff cancelling out, leaving just (3n+5)/(3n+1) = 1 + 4/(3n+1). Evaluating this at n=2023 yields 1 + 2/3035 = 1.000658979
Analytic Solution
