+79 Part a:
Find polynomial f(n) such that for all integers \(n \geq 1\), we have
\(3\left( 1\cdot2 + 2\cdot3 + \ldots + n(n+1) \right) = f(n)\)
Write f(n) as a polynomial with terms in descending order of n.
Part b:
The closed form sum of
\(12 \left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]\)
for \(n \geq 1\) is \(n(n+1)(n+2)(an+b)\) Find \(an + b.\)
+129852 First one
The successive sums are
6 , 24 , 60 , 120, 210
Sum of differences 18 36 60 90
18 24 30
6 6
We have 3 non-zero row differences.......so the polynomial will be of the form
an^3 + bn^2 + cn + d = f(n)
We can solve this system
a + b + c + d = 6
8a + 4b + 2c + d = 24
27a + 9b + 3c + d = 60
64a + 16b + 4c + d = 120
The solution is a = 1 b = 3 c = 2 and d = 0
So f(n) = n^3 + 3n^2 + 2n
