SOLUTION: The closed form sum of
12 [ 1^2 * 2 + 2^2 * 3 + ... + n^2 (n+1) ]
for n <= 1 is n(n+1)(n+2)(an+b). Find an + b.
Algebra.Com
Question 1177060: The closed form sum of
12 [ 1^2 * 2 + 2^2 * 3 + ... + n^2 (n+1) ]
for n <= 1 is n(n+1)(n+2)(an+b). Find an + b.
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
Obviously, from the form of the expression, the statement of the problem is supposed to say that the closed form of the sum is for n >=1 -- not for n <= 1.
There are only two unknowns in the problem -- a and b. So two equations in a and b should be enough to solve the problem.
n=1....
-->
n=2....
-->
a+b=4; 2a+b=7 --> a=3, b=1.
ANSWER: an+b = 3n+1
CHECK for n=3....
12(2(1^)+3(2^2)+4(3^2) = 12(2+12+36) = 600
(3)(4)(5)(3(3)+1) = 600
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