How do we find the term n^(th) and sum to n^(th) of the following series?
1)1*2^2+2*3^2+3*4^2+...n^(th) term
2)1^2+(1^2+2^2)+(1^2+2^2+3^2)+....n^(th)) term
2 Answers
Explanation:
I won't derive these common sums:
The sum of the squares is
The sum of cubes is remarkably the square of the triangular numbers, the sums of the natural numbers.
Now,
Is that a second sum? I'll leave that to others.
Second sum:
Explanation:
Let's do the second sum.
The
There are
The inner sum is known:
The sum of cubes is the square of the triangular number,
The triangular numbers are the sums of the naturals,