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avatar+645 

Find polynomial f(n) such that for all integers n >= 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 

 Feb 14, 2018
 #1
avatar+111331 
+1

We have  the first few terms as

 

       2      8      20     40     70   

            6     12      20     30

                6       8       10

                     2       2

 

The Sum of Differences  produces 3 non-zero rows...so we will have a  cubic polynomial  representing the sum inside the parentheses

 

So we have that

 

a  +  b +  c   +  d   =  2

8a + 4b + 2c + d  =  8

27a + 9b + 3c + d  =  20

64a + 16b + 4c + d  =  40

 

Solving this system produces    a  = 1/3   b  =  1   c  = 2/3    d  = 0

 

So...the polynomial generating  the sums inside the parentheses  is just

 

(1/3)n^3 +  n^2  +  (2/3)n

 

So  3 times this is just

 

f (n)  =  n^3  + n^2  + 2n

 

 

cool cool cool

 Feb 14, 2018
edited by CPhill  Feb 14, 2018
 #2
avatar+24995 
+1

Find polynomial f(n) such that for all integers \( n \ge 1\) we have
\(3\left( 1\cdot2 + 2\cdot3 + \ldots + n(n+1) \right) = f(n).\)

 

\(\begin{array}{|rcll|} \hline f(n) &=& 3\Big( 1\cdot2 + 2\cdot3 + 3\cdot4 + \ldots + n(n+1) \Big) \quad & | \quad n(n+1) = n^2+n \\ &=& 3 ( 1^2+1+2^2+2+3^2+3 + \ldots + n^2+n ) \\ &=& 3 ( \underbrace{1+2+3+ \ldots + n}_{=\dfrac{(n+1)n}{2}} + \underbrace{1^2 +2^2 +3^2 + \ldots + n^2}_{=\dfrac{(n+1)n(2n+1)}{6}} ) \\\\ &=& 3 \left( \dfrac{(n+1)n}{2} + \dfrac{(n+1)n(2n+1)}{6} \right) \\\\ &=& 3 \left[ \dfrac{(n+1)n}{2}\left(1+ \dfrac{2n+1}{3}\right) \right] \\\\ &=& \dfrac{(n+1)n}{2}\left(3+ 2n+1\right) \\\\ &=& \dfrac{(n+1)n}{2}\left(2n+4\right) \\\\ &=& (n+1)n(n+2) \\\\ &=& n(n+1)(n+2) \\\\ &\mathbf{=}& \mathbf{ n^3+3n^2+2n } \\\\ \hline \end{array}\)

 

laugh

 Feb 14, 2018

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