# General method for finding summation of series with $n^{th}$ term difference in AP

What is the method for finding the summation of a series whose $$n^{th}$$ difference between consecutive terms is in an AP?

For example, $$2,12,36,80,150,252...$$ Taking the first term difference we get another series as such - $$10,24,44,70,102...$$.

Taking it's term difference we get the series $$14,20,26,32$$ which is in an Arithmetic Progression.

Or, $$1,13,53,143,311,591,1023...$$

In this, the series formed by the $$3^{rd}$$ order difference is in AP.

• I don't have a proof for this, maybe other users can help, but one my teachers said that the general term of a series having $n^{th}$ order difference in AP is $a_1+a_2i+a_3i^2\cdots a_{n+1}i^n$. This is for the $i$th term. Notice that if you put $n=1$, you get the well known AP. Now, using $n+1$ terms of the sequence, find all $a_n$ and then sum each terms individually: $$S=a_1\Sigma 1+a_2\Sigma n\cdots$$ Jun 21, 2020 at 4:39
• @AryanSonwatikar You're basically correct. My answer to a related question of Doubt in finding general term of the given sequence includes stating more details are in Theorem $1$ of Difference Tables of Sequences. Jun 21, 2020 at 4:57
• @AryanSonwatikar okay, I understood what you are saying. I think this will do for what I was trying to do currently, thank you for your help. Just to be sure, is there any condition on when this is applicable? Jun 21, 2020 at 5:11

Let $$(a_k)_{k\geq 0}$$ be a sequence which is an arithmetical progression of order $$n$$. We show the following is valid: \begin{align*} \color{blue}{\sum_{k=0}^qa_k =\sum_{j=0}^n\binom{q+1}{j+1}\Delta^j a_0 \qquad\qquad q\geq n}\tag{0} \end{align*}

We use the shift operator $$E$$, which applied at $$a_k$$ gives $$E a_k=a_{k+1}$$, $$k\geq 0$$ and we als need the forward difference operator $$\Delta a_k=a_{k+1}-a_k$$. The shift operator $$E$$ can be written in terms of the $$\Delta$$ operator as \begin{align*} E&=\Delta+I\tag{1} \end{align*} since \begin{align*} E a_k&=a_{k+1}\\ (\Delta+I)a_k&=\Delta a_k + I a_k=(a_{k+1}-a_k)+a_k=a_{k+1}\\ \end{align*} where $$I a_k=a_k$$ is the identity operator.

Let's take a look at OP's example: \begin{align*} (a_k)_{k\geq 0}=(2,12,36,80,150,252,\ldots)\tag{2} \end{align*} which is an arithmetical progression of order $$3$$. To better see what's going on we calculate the differences up to the fourth order. \begin{align*} \begin{array}{r|ccccccccccccc} a_k\ &\color{blue}{2}&&12&&36&&80&&150&&252&&\cdots\\ \Delta a_k\ &&\color{blue}{10}&&24&&44&&70&&102&&\cdots\\ \Delta^2 a_k\ &&&\color{blue}{14}&&20&&26&&32&&\cdots\\ \Delta^3 a_k\ &&&&\color{blue}{6}&&6&&6&&\cdots\\ \Delta^4 a_k\ &&&&&0&&0&&\cdots\\ \end{array} \end{align*}

We see the row $$\Delta^3 a_k$$ contains the constant $$6$$ only, indicating $$(a_k)_{k\geq 0}$$ is an arithmetical progression of order $$3$$. The left-most elements of $$\Delta a_k^j, 0\leq j\leq 3$$ are marked in blue. These are coefficients which will be relevant for the wanted formula.

We can write $$E^m a_k=a_{k+m}$$ with $$m,k\geq 0$$ in terms of the difference operator and obtain applying the binomial theorem \begin{align*} E^m&=(\Delta+I)^m=\sum_{j=0}^m\binom{m}{j}\Delta^j I^{m-j}\\ E^m a_k=a_{k+m}&=(\Delta+I)^ma_k =\sum_{j=0}^m\binom{m}{j}\Delta^j I^{m-j}a_k\\ &=\sum_{j=0}^m\binom{m}{j}\Delta^j a_k\tag{3}\\ \end{align*} Setting $$k=0$$ and $$m=k$$ in (3) we obtain a representation of $$a_k$$ in terms of the $$j$$-th differences $$\Delta^j a_0$$ of $$a_0$$ \begin{align*} a_k=\sum_{j=0}^{\min\{k,n\}}\binom{k}{j}\Delta^j a_0\qquad\qquad k\geq 0\tag{4} \end{align*} Note there are no more than $$n+1$$ terms in the right-hand side of (3), since $$\Delta^j a_0=0$$ if $$j>n$$.

We finally calculate a formula for the sum of the first $$q$$ terms $$a_k$$ from (4). To simplify things somewhat, we only consider the case $$q\geq n$$.

We obtain for $$q\geq n$$: \begin{align*} \color{blue}{\sum_{k=0}^q}\color{blue}{a_k}&=\sum_{k=0}^q \sum_{j=0}^n\binom{k}{j}\Delta^j a_0\tag{5}\\ &=\sum_{j=0}^n\sum_{k=0}^q \binom{k}{j}\Delta^j a_0\tag{6}\\ &=\sum_{j=0}^n\Delta^j a_0\sum_{k=j}^q \binom{k}{j}\tag{7}\\ &\,\,\color{blue}{=\sum_{j=0}^n\binom{q+1}{j+1}\Delta^j a_0}\tag{8}\\ \end{align*} and the claim (0) follows.

Comment:

• In (5) we use formula (4).

• In (6) we exchange the sums.

• In (7) we factor out $$\Delta^j a_0$$ and set the lower index of the inner sum to $$j$$ since $$\binom{k}{j}=0$$ if $$0\leq k.

• In (8) we apply the Hockey stick identity.

Now its time to harvest. We apply formula (8) to calculate for instance in OP's example the sum of the first $$q+1=6$$ elements. On the one hand we have from (2) \begin{align*} \sum_{k=0}^5a_k&=2+12+36+80+150+252\\ &\,\,\color{blue}{=532} \end{align*} on the other hand we obtain from (8) and the blue marked entries from the differences triangle: \begin{align*} \sum_{k=0}^5a_k&=\sum_{j=0}^3\binom{6}{j+1}\Delta^j a_0\\ &=\binom{6}{1}2+\binom{6}{2}10+\binom{6}{3}14+\binom{6}{4}6\\ &=12+150+280+90\\ &\,\,\color{blue}{=532} \end{align*}

Note that in the first calculation we always have to add $$q+1$$ terms in case of $$q\geq n$$, while in the second we have to sum up no more than $$n+1$$ terms, $$n$$ giving the order of the arithmetic progression.

Note: As preparatory work for this answer I went through section 23: Expansion into a Series of Binomial Coefficients of Calculus of finite Differences by C. Jordan.

Let $$\sum_{i = 1}^{n} a_{i}$$ be the series and let $$b_{n} = a_{n + 1} - a_{n}$$ be the sequence of differences of $$a_{n}$$. If the first term of the sequence is $$a_{1}$$ and $$b_{n}$$ is an arithmetic sequence with common difference $$d$$, then the formula for the sum of the series in terms of $$a_{1}$$, $$b_{1}$$ and $$d$$ is

$$\sum_{i = 1}^{n} a_{i} = na_{1} + n(n - 1)b_{1} + \frac{n(n - 1)(2n - 1)}{6}d$$

Proof

Since $$b_{n}$$ is an arithmetic progression, then each term $$b_{i}$$ can be expressed in terms of its first term $$b_{1}$$ and common difference $$d$$ as

$$b_{i} = b_{1} + (n - 1)d$$

By definition of $$b_{n}$$, we have

$$a_{i + 1} - a_{i} = b_{1} + (n - 1)d$$

Summing up all sides up to $$n - 1$$, we have

$$\sum_{i = 1}^{n - 1} a_{i + 1} - \sum_{i = 1}^{n - 1} a_{i} = (n - 1)b_{1} + {(n - 1)}^2d$$

The terms cancel in the left-hand side so that $$a_{n} - a_{1}$$ remains. Transposing $$a_{1}$$ to the right-hand side, we have

$$a_{n} = a_{1} + (n - 1)b_{1} + + {(n - 1)}^2d$$

At this point, we have derived the general term of the series $$\sum_{i = 1}^{n} a_{i}$$ in terms of $$a_{1}$$, $$b_{1}$$ and $$d$$. Summing up both sides up to $$n$$, we have

$$\sum_{i = 1}^{n} a_{i} = na_{1} + n(n - 1)b_{1} + \left(\sum_{i = 1}^{n} {(n - 1)}^2\right)d$$

From the sum of squares formula, we have

$$\sum_{i = 1}^{n} (n - 1)^2 = \frac{n(n - 1)(2n - 1)}{6}$$

Plugging this into the last result, we have

$$\sum_{i = 1}^{n} a_{i} = na_{1} + n(n - 1)b_{1} + \frac{n(n - 1)(2n - 1)}{6}d$$

and the proof is done.

• Thanks, although I was asking if there was a general way to compute the sum for any order of difference. From what I understand, this is the formula when the first order difference is in AP. @AryanSonwatikar has shared one but it is without a proof. Jun 21, 2020 at 6:47