1+2+3+...+(4^2- 1^2- 3) / 2 , the left hand side of this equation is a telescoping sum, namely every time we add a term of the same for part of the new term cancels out part of the existing sum.
1. find the corresponding sum if we add up the first N non-negative integers. this should simplify our formula. 2. adapt this method to find the sum of the squares of the first N -non-negative integers. to obtain the telescoping component, use (m+1)^3 -m^3=3m^2+3m+1 3. this method can be adapted to find the sum of the Kth powers of the first N non-negative integers and the sum of the fourth powers of the first N non-negative integers.
4. find a formula for the sum of the 10th powers of the first N non-negative integers that involves a close form featuring an algebraic combination including N.
help would be appreciated, please number the responses!!