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.

  • $\begingroup$ 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. $\endgroup$ – 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 $a_{2}=k\cdot 1^3+p\cdot 1^2+w\cdot 1+q=k+p+w+q=6$

  • $\begingroup$ 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. $\endgroup$ – David K Feb 8 '17 at 2:31
  • $\begingroup$ @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. :) $\endgroup$ – kingW3 Feb 8 '17 at 2:45
  • $\begingroup$ I think that's likely how you're supposed to do this problem. My way is much less economical. $\endgroup$ – David K Feb 8 '17 at 3:55
  • $\begingroup$ Is there a way to find out what the values of k, p, w and q are individually? $\endgroup$ – user3753 Feb 8 '17 at 3:56
  • $\begingroup$ @user3753 Plug in $n=2,3,4$ then you have 4 equations with 4 unknowns. $\endgroup$ – 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 = $1+3+2=\color{red}6$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.