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(Part 1) Find polynomial f(n) such that for all integers n>=1, we have 3(1*2+2*3+...+n(n+1))=f(n). Write f(n) as a polynomial with terms in descending order or n.

 

(Part 2) 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.

Guest May 23, 2018
 #1
avatar+20024 
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(Part 1)

Find polynomial f(n) such that for all integers n>=1, we have 3(1*2+2*3+...+n(n+1))=f(n).

Write f(n) as a polynomial with terms in descending order of n.

 

\(\begin{array}{|rcll|} \hline f(n) &=& 3\cdot[~ 1\cdot 2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 + \dots +n(n+1)~] \\ f(n) &=& 3\cdot[~ 1\cdot (1+1) + 2\cdot (2+1) + 3\cdot (3+1) + 4\cdot (4+1) + \dots +n(n+1)~] \\ f(n) &=& 3\cdot[~ (1^2+1) + (2^2+2) + (3^3+3) + (4^2+4) + \dots +(n^2+n)~] \\ f(n) &=& 3\cdot[~ (1+2+3+4+ \dots +n) + (1^2+2^2+3^3+4^2 + \dots +n^2)~] \\\\ && (1+2+3+4+ \dots +n) = \frac{(1+n)\cdot n}{2}\\ && (1^2+2^2+3^3+4^2 + \dots +n^2)= \frac{n\cdot(n+1)\cdot(2n+1)}{6} \\\\ f(n) &=& 3\cdot[~ \frac{(1+n)\cdot n}{2} + \frac{n\cdot(n+1)\cdot(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot[~ \frac{1}{2} + \frac{(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot[~ \frac{3}{6} + \frac{(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot( \frac{3+2n+1}{6} ) \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot( \frac{2n+4}{6} ) \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot 2\cdot( \frac{n+2}{6} ) \\ \mathbf{f(n) }&\mathbf{=} &\mathbf{ n \cdot (n+1)\cdot (n+2)} \\ \hline \end{array}\)

 

Terms in descending order of n

\(\boxed{f(n) = n^3+3n^2+2n}\)

 

laugh

heureka  May 24, 2018
edited by heureka  May 24, 2018
 #2
avatar+20024 
0

(Part 2)

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.

 

\(\begin{array}{|rcll|} \hline && \mathbf{12 \left[1^2\cdot 2 + 2^2\cdot 3 + 3^2\cdot 4 +\ldots +n^2 \cdot (n+1)\right] } \\ &=& 12\sum \limits_{k=1}^{n} k^2 \cdot (k+1) \\ &=& 12\sum \limits_{k=1}^{n} (k^3 + k^2) \\ &=& 12\left[\sum \limits_{k=1}^{n} (k^3) + \sum \limits_{k=1}^{n} (k^2) \right] \\ && \begin{array}{|rcll|} \hline \sum \limits_{k=1}^{n} (k^3) &=& 1^3+2^3+3^3+ \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\ \sum \limits_{k=1}^{n} (k^2) &=& 1^2+2^2+3^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \hline \end{array} \\ &=& 12\left[ \left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6} \right] \\ &=& 12n(n+1)\left[ \frac{n(n+1)}{4} + \frac{2n+1}{6} \right] \\ &=& 12n(n+1)\left[ \frac{n(n+1)}{4} + \frac{2n+1}{2\cdot 3} \right] \\ &=& 12n(n+1)\left[ \frac{n(n+1)}{4} + \frac{n+\frac12}{3} \right] \\ &=& 12n(n+1)\left[ \frac{3n(n+1)+4n+2}{12} \right] \\ &=& n(n+1)[3n(n+1)+4n+2] \\ &=& n(n+1)[3n(n+2)-3n+4n+2] \\ &=& n(n+1)[3n(n+2) + n+2] \\ &\mathbf{=}& \mathbf{ n(n+1)(n+2)(3n+1) } \\ \hline \end{array} \)

 

\(\mathbf{\large{an+b = 3n+1}}\)

 

laugh

heureka  May 24, 2018

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