Sum of a Series whose difference is an Arithmetic Progression

Question

Problem Statement

Find the sum of $T_n$ and $S_n$ where $T_i$ denotes the $i^{th}$ term and $S_i$ denotes the sum of the first i terms for the following series : $$ 3, 1, 31,69,131,223 ... $$

I am not in search of solution to this problem, since the general solution is as explained here in Sum of terms whose differences are in Arithmetic progression..

More specifically speaking,

If the 2nd difference is constant, we have a quadratic sequence. Meaning, the $n$th term is $an^2+bn+c$ for some $a,b,c$

But we could model this quadratic as Binomial polynomial too as explained in this wikipedia article and,

Over any field of characteristic $0$ (that is, any field that contains the rational numbers), each polynomial $p(t)$ of degree at most $d$ is uniquely expressible as a linear combination $\sum_{k=0}^{d} a_k \binom{t}{k}$ of binomial coefficients, because the binomial coefficients consist of one polynomial of each degree. The coefficient $a_k$ is the $k^{th}$ difference of the sequence $p(0), p(1), ..., p(k)$

This whole thing just escapes me. Even the resource where I saw this problem has used this approach and I dont seem to be able to make sense of it. Why can we do this? More importantly, how does this help us?

Answer 1

Let's look at quadratics, so, degree $d=2$. It should be clear that any linear combination $\sum_{k=0}^2c_k{t\choose k}$ is a quadratic: $$ \sum_{k=0}^2c_k{t\choose k}=c_0{t\choose0}+c_1{t\choose1}+c_2{t\choose2}=c_0+c_1t+c_2{t(t-1)\over2}=e+ft+gt^2 $$ where $e=c_0$, $f=c_1-(1/2)c_2$, and $g=(1/2)c_2$. Going in the other direction, any quadratic $e+ft+gt^2$ can be expressed as $c_0{t\choose0}+c_1{t\choose1}+c_2{t\choose2}$ where $c_0=e$, $c_1=f+g$, and $c_2=2g$.

Finally, given the quadratic $p(t)=e+ft+gt^2$, we find $$ p(0)=e,\qquad p(1)=e+f+g,\qquad p(2)=e+2f+4g $$ with first differences $$ f+g,\qquad f+3g $$ and second difference $$ 2g $$ The leftmost entry in these three displays gives us $e,f+g,2g$, which is exactly what we found as the values of $c_0,c_1,c_2$ in terms of $e,f,g$, and that demonstrates the bit about the coefficients (in the linear combination) coming from the differences of the sequence of values of the polynomial $p(t)$.

Answer 2

Why can we do this? What about using such a polynomial for finding out the sum of the series and the nth term of the series?

Presumably you already know how to calculate the sum of an arithmetic series: $$\text{sum}=(\text{number of terms}) \times \frac{(\text{first term}+\text{last term})}{2}.$$

Consider $(u_n)$ an arithmetic sequence with constant difference $d$, and define its series $(s_n) = \left( \sum_{k=0}^n u_k \right)$.

Then: \begin{align*} u_n & = u_0 + n d \\ s_n & = (n+1)\frac{u_0+u_n}{2} \\ & = u_0 + (u_0+\frac{d}{2})n + \frac{d}{2}n^2 \end{align*}

So the $n$th term of an arithmetic sequence is a degree-1 polynomial, and the $n$th partial sum of the series is a degree-2 polynomial.

If you sum once more, i.e. take the series of the partial sums of the series, then you'll end up with a degree-3 polynomial, and so on.

Knowing this, if you know $s_0,s_1,s_2$ but don't know $d$ and $u_n$ and $s_n$, you can find the formula for $s_n$ by writing $s_n = an^2+bn+c$ and solving the system of three equations in unknowns $a,b,c$.

Answer 3

Generating functions can also be useful.

Note: This is one of my "nothing original" posts.

In this case, if $s_n =\sum_{k=n0}^n r_k $, $S(x) =\sum_{n=n0}^{\infty} s_nx^n $ and $R(x) =\sum_{n=n0}^{\infty} r_nx^n $

then $S(x) =\dfrac{R(x)}{1-x} $.

(I start the terms at $n0$ so it works no matter where the series start.)

The proof is standard, just reversing the order of summation.

I am not considering convergence.

$\begin{array}\\ S(x) &=\sum_{n=n0}^{\infty} s_nx^n\\ &=\sum_{n=n0}^{\infty} x^n\sum_{k=n0}^n r_k\\ &=\sum_{k=n0}^{\infty}\sum_{n=k}^{\infty} x^n r_k\\ &=\sum_{k=n0}^{\infty}r_k\sum_{n=k}^{\infty} x^n \\ &=\sum_{k=n0}^{\infty}r_kx^k\sum_{n=k}^{\infty} x^{n-k} \\ &=\sum_{k=n0}^{\infty}r_kx^k\sum_{n=0}^{\infty} x^{n} \\ &=\sum_{k=n0}^{\infty}r_k\dfrac{x^k}{1-x}\\ &=\dfrac1{1-x}\sum_{k=n0}^{\infty}r_kx^k\\ &=\dfrac{R(x)}{1-x}\\ \end{array} $