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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 $s_n = an^3 + ... 1 As has been pointed out, there is no foolproof way to solve these problems, so it is best (in general) to rely on intuition. We can devise a reasonable pattern for just about anything, demonstrating why all-purpose techniques can never work. For example, given any 6th term in your sequence, we can use polynomial interpolation to fit a polynomial of degree ... 2 To reiterate what has been said in the comments, there is no surefire way to solve "any kind of series problem." Suppose you are given$n$digits with an obvious pattern--the$n+1th 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 ... 0 Yes, there is: \begin{align*} \frac{2(2n-1)}{n+1}C_{n-1}&=\frac{2(2n-1)}{n(n+1)}\binom{2n-2}{n-1}\\\\ &=\frac2{n+1}\binom{2n-1}n\\\\ &=\frac1{n+1}\left(\binom{2n-1}n+\binom{2n-1}n\right)\\\\ &=\frac1{n+1}\left(\binom{2n-1}n+\binom{2n-1}{n+1}\right)\\\\ &=\frac1{n+1}\binom{2n}n\\\\ &=C_n\;. \end{align*} 0 Sure, $$C_{n+1} = \frac{2(2n+1)}{n+2}C_n.$$ You can prove this from the formula $$C_n = \frac{1}{n+1}\binom{2n}{n}.$$ More at Wikipedia: https://en.wikipedia.org/wiki/Catalan_number. 2 One possible answer: The first term is greater than the second term by the third term raised to itself. Using a formula, the tuples are generated by $$(x+y^y,x,y)$$ for some givenx$and$y$. It is easy to check that the two given tuples follow this rule. That would make the third tuple $$(12+4^4,12,4)$$ or$?=268\$. Note that this is not the only possible ...