Sum Of The Coefficients Of The n+1 Term

The 1st term of a sequence of positive integers is $2$; the second term is $6$; the third term is $12$; and the fourth term is $20$. The sequence continues in this manner with the positive difference between successive terms increasing by $2$ each time.The $(n+1)$ term of these sequence can be expressed as $kn^3+pn^2+wn+q$ where $k,p,w,q$ are integers.

How could I find the sum of $k,p,w,q$? I know for a fact that the answer is 6. However I am unsure of the process.

• If you want to know the individual coefficients of $kn^3+pn^2+wn+q$ (although you really don't need them for this question), one way to find them is the method of differences. See math.stackexchange.com/questions/999324/… for example. – David K Feb 10 '17 at 4:00

You have that $a_{n+1}=kn^3+pn^2+wn+q$ then plugging in $n=1$ you have that $a_{2}=k\cdot 1^3+p\cdot 1^2+w\cdot 1+q=k+p+w+q=6$

• Wow, that way actually requires some insightful thinking. I just recognized the formula for $a_{n+1}$ as a product of linear factors and multiplied them to get the polynomial. – David K Feb 8 '17 at 2:31
• @DavidK I couldn't quite recognize the formula so I thought I should create a system of equations, I started by plugging $n=1$ and voila. :) – kingW3 Feb 8 '17 at 2:45
• I think that's likely how you're supposed to do this problem. My way is much less economical. – David K Feb 8 '17 at 3:55
• Is there a way to find out what the values of k, p, w and q are individually? – user3753 Feb 8 '17 at 3:56
• @user3753 Plug in $n=2,3,4$ then you have 4 equations with 4 unknowns. – kingW3 Feb 8 '17 at 15:40

The shortest approach (if for an exam) is that given in the solution posted by @kingW3.

For additional information, the following might be useful.

\begin{align} \big\lbrace 2,6,12,20,...\big\rbrace &=2\cdot \big\lbrace 1,3,6,10,...\big\rbrace\\ &=2\cdot \binom n2 &&(n=1,2,3,4,...)\\ &=n(n+1) \end{align}

The $(n+1)$-th term is $(n+1)(n+2)=n^2+3n+2$. Sum of coefficients = $1+3+2=\color{red}6$.