How do I solve the missing number in a series using the application of calculus?

Question

5,6,10,19,35, ...

I know the way to solve this with the basic way i.e. finding the difference. Otherwise, look for the square and difference since the number is growing kind of exponentially. However, I am looking for a sure shot method of solving any series questions. This should be using calculus. I want to make a perfect way to solve any kind of series problem. My question is how can I solve such questions with Calculus.

Answer 1

To reiterate what has been said in the comments, there is no surefire way to solve "any kind of series problem." Suppose you are given $n$ digits with an obvious pattern--the $n+1$th digit could be anything. You always need to be given additional information when just handed a string of numbers, even if the pattern is as obvious as Andre's comment points out.

Answer 2

As has been pointed out, there is no foolproof way to solve these problems, so it is best (in general) to rely on intuition. We can devise a reasonable pattern for just about anything, demonstrating why all-purpose techniques can never work.

For example, given any 6th term in your sequence, we can use polynomial interpolation to fit a polynomial of degree 5 to the points $(n, S_n)$ where $S_n$ is the $n$th term, and so on for an arbitrary number of points.

Edit: Since I brought it up, I may as well note that polynomial interpolation does give what is probably the most intuitive description of your series (albeit not in the most intuitive form). Fitting a 4th-degree polynomial to your series, we find $\frac{1}{3}x^3-\frac{1}{2}x^2+\frac{1}{6}x+5$. (The $x^4$ coefficient is 0 $-$ a good sign, as it implies that the last term in the sequence "agrees with" the polynomial growth pattern established by the first three without further coercion.) Indeed, this sequence increases term-to-term by successive squares.

Answer 3

You can look at the consecutive differences until you reach a constant difference. $$5,\; 6,\; 10, \;19,\; 35$$ $$1,\; 4, \;9,\; 16$$ $$3,\; 5,\; 7$$ $$2, \;2.$$ Using the fact that the $n$-th differences of a sequence $\{s_n\}$ are constant and non-zero if and only if $\{s_n\}$ is generated by an $n$-th degree polynomial, we can conclude that $s_n = an^3 + bn^2 + cn + d$. We immediately have $d = s_0 = 5$. Now use the fact that if a sequence is generated by an $n$-th degree polynomial with leading coefficient $c_n$, the $n$-th differences are constantly $(c_n)n!$. In our case, $3!\,a = 2$, so $a = \frac{1}{3}.$ To solve for $b$ and $c$, we use the initial data. $$s_1 = a + b + c + d = 6 \Rightarrow b + c = \frac{2}{3}$$ $$s_2 = 8a + 4b + 2c + d = 10 \Rightarrow 4b + 2c = \frac{7}{3}.$$ Some algebra later, we have $b = \frac{1}{2}$ and $c = \frac{1}{6}$. Thus the $(n+1)$-th term in the sequence is given by $$s_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n + 5,$$ and the $n$-th term is given by \begin{align*} s'_n &= \frac{1}{3}(n-1)^3 + \frac{1}{2}(n-1)^2 + \frac{1}{6}(n-1) + 5 \\ &= \frac{1}{3}n^3 - \frac{1}{2}n^2 + \frac{1}{6}n + 5. \end{align*} This approach doesn't use standard calculus, but rather calculus of differences. Some of the analogs are pretty clear: the $n$-th differences resemble $n$-th derivatives, in the sense that both are computed by subtracting "close-by" terms and that the $n$-th difference/derivative is constant and equal to $a\, n!$. This is a much more elegant solution than trying to interpolate a 3rd degree polynomial much more generally (though some may argue it's effectively the same by forcing the $y$-intercept to be the constant term and using differentiation to find the leading coefficient).

N.B. The commenters on your post are perfectly correct in that this sequence really could be anything (who knows, maybe the next several terms are all 0). The approach I have chosen begins with the hunch that we will eventually reach constant differences and that the constant differences are the same throughout the entire sequence.