4

For the series, $3, 7, 14, 24, 37, \ldots$, the $1$st successive differences are $4,7,10,13,\ldots$, and the $2$nd successive differences are $3,3,3,\ldots$. So, the book says, the $nth$ term $T_n$ of the given series will be $an^2+bn+c$.

And for the series, $3,8,22,72,266,1036,\ldots$, the $1$st successive differences are $5,14,50,194,770,\ldots$, and the $2$nd successive differences are $9,36,144,576,\ldots$, which are in geometric progression with common ratio being $4$. So, the book says $T_n$ will be $a4^{n-1}+bn+c$.

I have verified in both cases that it's true with $a,b,c$ in first case coming out to be $\frac32,-\frac12,2$, and in second case $1,2,0$, respectively.

My question is why is this so? Why is $T_n$ the way it is? How to approach this method in a fresh question?

Blue
  • 75,673
aarbee
  • 8,246
  • 3
    Taking successive differences is very close to differentiating. Think about that, and the answer may appear obvious. – Arthur Jul 26 '19 at 10:59

1 Answers1

3

Call the first and second sequences $a_n,\,b_n$, and for any sequence $u_n$ define $\Delta u_n:=u_{n+1}-u_n$, so we can prove the following results (using early ones in the inductive step of proving later ones): $$\Delta^2a_n=3,\,\Delta a_n=3n+1,\,a_n=\frac32 n^2-\frac12 n+2,\\\Delta^2 b_n= 9\times 4^{n-1},\,\Delta b_n=3\times 4^{n-1}+2,\,b_n=4^{n-1}+2n.$$

J.G.
  • 115,835
  • Integrating $3$, we get, $3n+k$. By putting values, we get $k$ as $1$. Integrating $3n+1$, we get, $\frac32n^2+n+k$. Sorry, why have you written $-\frac12n$? Integrating $94^{n-1}$ gives us $94^{n-1}/ln4$. Perhaps this is not what you are doing. – aarbee Jul 26 '19 at 11:31
  • 1
    @Ramit Integration gives you a rough idea, but it's inexact for inferring sequences, except when $\Delta u_n$ is constant. As I said, you need to use induction. – J.G. Jul 26 '19 at 11:41
  • Induction, the way I understand, is a way where we prove a result for n=1. Then we assume it to be true for n=k. Then we prove it for n=k+1. I am not sure how to apply that here. – aarbee Jul 26 '19 at 15:38
  • @Ramit Example: since $\Delta^2 a_n= 3$, $\Delta a_k=3k+1\implies\Delta a_{k+1}=3(k+1)=1$, so the behaviour of $\Delta a_n$ is established. Now keep going. – J.G. Jul 26 '19 at 15:45