I have this series comprised of


and so on. So far i have found that they add up in intervals being odd numbers. But I don't know how to find, let's say the 16th term and the sum up to that term. What should I do?


Although it is somewhat possible to calculate the sum up to the 16th term in the above equation, what if the series is of higher order like this?.


Is there a shortcut to calculate the sum up to the 16th term? Can this be done by hand or would be necessary to use software like Maple?.

  • 1
    $\begingroup$ So, we have $a_{n+1}-a_n=2n-1$ with $a_1=1$. Can you solve this recurrence relation? $\endgroup$ – Prasun Biswas Sep 30 '17 at 8:31

While Opt's answer is correct, it can take a good deal of practice to come up with such accurate observations on an unknown series. A very good and straightforward method in such a case is the method of differences

It formally states:

any term, of a given number series, can be expressed as a polynomial of degree $n$, if the $n$ differences of the series is constant.

Let me illustrate with your series.


The first differences are $$1(=2-1);3(=5-2); 5(=10-5);7(=17-10);9(=26-17)$$.

The second differences are $$2(=3-1); 2(=5-3); 2(=7-5);2(=9-7)$$ Note that all the second differences are constant! This implies, by the method of differences statement, that the $r$-th term in your series can be represented as a quadratic polynomial (degree 2) i.e. $T_r = ar^2+br+c$ for some constants $a,b,c$

Now, all you need to do is equate $T_1 =1; T_2=2; T_3=5$ and solve for $a,b,c$.

Added on request: To solve the above, you only need to do write them out after putting in $n=1,2,3$:

$$a+b+c=1...E1$$ $$4a+2b+c=2...E2$$ $$9a+3b+c=5...E3$$

You have three variables and three equations, thus they can be easily solved by subtraction and elimination. $E2-E1$ and $E3-E2$ respectively gives two new sets of equations:

$$3a+b=1...E4$$ $$5a+b=3...E5$$

and then $E5-E4$ straightaway gives $a=1$. Putting this in E5 gives $b=-2$. Putting them in $E1$ gives $c=2$.
Thus, our $T_r=r^2-2r+2=1+(r-1)^2$

For summation of terms from 1 to 16, you need to do:


for which you may easily use the summation identities for $r^2$ and $r$.

Hope that helps!

  • $\begingroup$ Sorry but my level of algebra isn't very proficient to solve the latter part of your explanation, mind if you extend a bit?. According to the method you proposed the polynomial is in terms of x but how does this translates as 'nth' term?. I do understand the nth is the degree but what's the x?. How do I solve for a, b and c, Can you show it to me please?. $\endgroup$ – Chris Steinbeck Bell Sep 30 '17 at 9:35
  • 2
    $\begingroup$ For the general case (when the $n$-th difference is constant), it is simpler to use the binomial polynomials $\dbinom xi$ as a basis for the vector space of polynomials. $\endgroup$ – Bernard Sep 30 '17 at 9:37
  • $\begingroup$ @Bernard Can you be a bit explicit on how do the binomial coefficient replaces constants in the above equation?. Am I understanding correctly?. Gaurang Tandon, There is also missing the last part of my question which involves the sum from 1 to the 16th term in the series, how can I do that?. $\endgroup$ – Chris Steinbeck Bell Sep 30 '17 at 9:46
  • $\begingroup$ I'm not sure I was clear enough: I meant the canonical basis $1,x,x^2,\dots$ is replaced with the basis $\dbinom x0,\dbinom x1, \dbinom x2, \dots$. It's the natural basis in this situation because $\Delta\dbinom xk=\dbinom x{k-1}$. $\endgroup$ – Bernard Sep 30 '17 at 9:50
  • $\begingroup$ @ChrisSteinbeckBell That confusion about $n$ and $x$ was my mistake. Sorry for that. Now, I've added extensive elaboration. Hope that helps! $\endgroup$ – Gaurang Tandon Sep 30 '17 at 9:56

As Guarang Tandon says. You can create a table of differences between consecutive terms and then repeat that process on the list you just created.

\begin{array}{|l|c|ccccccc|} \hline \text{index} & n & 0 & 1 & 2 & 3 & 4 & 5 & \dots\\ \hline \text{sequence} & f_n & 1 & 2 & 5 & 10 & 17 & 26 & \dots \\ \text{first differences} & \Delta f_n && 1 & 3 & 5 & 7 & 9 & \dots \\ \text{second differences} & \Delta^2 f_n &&& 2 & 2 & 2 & 2 & \dots \\ \hline \end{array}

If you are lucky, at some point, all of the differences will be constant. (Of course this is an unverifiable claim. All we know for sure is that the first four second differences are constant. We can only assume that this fortuitous pattern continues.)

It can be proved that, if $\Delta^d f_n$ is constant, then then $f_n$ is a $d$th degree polynomial in $n$.

If the second differences are constant, then the first differences must be an arithmetic sequence. In this case, we see that $\Delta f_n = 2n+1$. It follows that $$f_{n+1} = f_n + 2n+1 \tag 1$$.

If the first differences are an arithmetic sequence, then $f_n = an^2 + bn + c$ for some real numbers $a, b,$ and $c$.

We can now use modified form of mathematical induction to find the values of $a,b, $ and $c$.

Since $f_0=1$, then $1 = a\cdot0^2 + b\cdot 0 + c = c$. So $f_n=an^2+bn+1$.

First we note that $$f_{n+1} = a(n+1)^2 + b(n+1) + 1 = f_n + 2an + (a+b).$$ Comparing that to $f_{n+1}= f_n+ 2n+1$, we see that $2a=2$ and $a+b=1$. Hence $a=1$ and $b=0$.

We conclude that $f_n=n^2+1$.

We know that $\sum_{k=0}^n k^2=\frac 16n(n+1)(2n+1)$.

Then \begin{align} \sum_{k=0}^n f_k &= \sum_{k=0}^n (k^2+1) \\ &= \sum_{k=0}^n k^2 + \sum_{k=0}^n 1 \\ &= \frac 16n(n+1)(2n+1) + (n+1) \\ &= \frac 16(n+1)(2n^2+n) + \frac 16(n+1)6 \\ &= \frac 16(n+1)(2n^2+n+6) \\ \end{align}


$\qquad \sum_{k=0}^5 f_k = 1+2+5+10+17+26=61$

$\qquad \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=5}=\frac 16(6)(61)=61$

So, since we started counting at $n=0$, the $16$th term is $f_{15} = 15^2+1=226$

and the sum of the first $16$ terms is $\sum_{k=0}^{15} f_k = \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=15}=1256$.

  • 2
    $\begingroup$ That's a nice table! :D $\endgroup$ – Gaurang Tandon Oct 1 '17 at 2:52

That's $$ 1 + 0 \\ 1 + 1 \\ 1 + 4 \\ 1 +9\\ 1 +16\\ 1 +25 \\ \vdots $$ Therefore, $$a_n = 1+(n-1)^2,$$

implying that $a_{16} = 1+(15)^2 = 226$.


As is said in other answers, the method of differences is certainly a way to go. However, in the simultaneous equations, you can obtain $a$ immediately:

If the $d^{th}$ row of the difference table is some constant $q$, the coefficient of the leading term of the polynomial is $\dfrac{q}{d!}$.

So, with the difference table as calculated in other answers, we have the second row is a constant 2. This gives us $a=\dfrac{2}{2!}=1$ without resorting to simultaneous equations.

Now, if you wanted to, you could use simultaneous equations to determine the other coefficients or you could subtract $n^2$ from each term:

$$1-0^2, 2-1^2, 5-2^2, 10-3^2, 17-4^2, 26-5^2,...\\ \text{which is }1,1,1,1,1,...$$

From which, the polynomial is $\fbox{n²+1}$


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.