The sum of the squares of the first n positive integers is n cube/3 + n square/2 + n/6 . If the sum of the squares of the first n odd positive integers is an3 + bn2 + cn + d, then find the value of (3a – 6c).
Answers
Answer:
01
Step-by-step explanation:
As usual, the first n in the table is zero, which isn't a natural number.
Because Δ3 is a constant, the sum is a cubic of the form
an3+bn2+cn+d, [1.0]
Answer:
the sum of square numbers? The sum of the first n squares, 12+22+... + n 2 = n ( n +1)(2 n +1)/6.
The formula says that the sum of the first n terms of our arithmetic sequence is equal to n divided by 2 times the sum of twice the beginning term, a, and the product of d, the common difference, and n minus 1. The n stands for the number of terms we are adding together.Sum of the First n Natural Numbers
We prove the formula 1+ 2+ ... + n = n(n+1) / 2, for n a natural number. There is a simple applet showing the essence of the inductive proof of this result.