Asked by warnar
1.2.3+2.3.5+3.4.7+...to n terms
Answers
Answered by
Steve
again with this stuff?
n
∑k(k+1)(2k+1)
k=1
= ∑(2k^3+3k^2+k)
= 2∑k^2 + 3∑k^2 + ∑k
Now just consult your reference for special sums of powers to get a 4th-degree polynomial for n.
n
∑k(k+1)(2k+1)
k=1
= ∑(2k^3+3k^2+k)
= 2∑k^2 + 3∑k^2 + ∑k
Now just consult your reference for special sums of powers to get a 4th-degree polynomial for n.
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