The \[{20^{th}}\] terms of the series \[2 + 3 + 5 + 9 + 16{\rm{ }} + .......\] is
A. \[950\]
B. \[975\]
C. \[990\]
D. \[1010\]

Hint: We discover the common distinction between the Arithmetic progression and the series. Then we take a few terms from each series to locate two common terms that will form our new common term series. Using the two terms in the new series, we determine the common variance of the new series and its total using the sum of n terms formula.

Formula Used: Arithmetic progression can be calculated by
\[{a_n} = a + (n - 1)d\]
N term’s sum can be determined by
\[{S_n} = \dfrac{n}{2}(2a + (n - 1)d)\]

Complete step by step solution: We have been provided in the question that,
The series is,
\[2 + 3 + 5 + 9 + 16{\rm{ }} + .......\]
And we are asked to determine the \[{20^{th}}\] term
Let us consider the equations of \[{S_1}\;\] as,
\[{S_1}\; = 2 + 3 + 5 + 9 + 16 + \ldots \ldots \ldots \ldots \ldots {x_n}\;\]------ (1)
\[{S_1}\; = {\rm{ }}2 + 3 + 5 + 9 + 16 + \ldots ..{x_{n - 1}}\; + {\rm{ }}{x_n}\;\]-------- (2)
Now, we have to subtract equation (1) and (2), we have
We retain the first term in (1) and remove the first term in (2) from the second term in (1) and so on
\[0{\rm{ }} = {\rm{ }}2 + \left[ {1 + {\rm{ }}2{\rm{ }} + {\rm{ }}4{\rm{ }} + {\rm{ }}7{\rm{ }} + {\rm{ }} \ldots to{\rm{ }}\left( {n - 1} \right){\rm{ }}terms} \right]{\rm{ }}-{\rm{ }}{x_n}\]
\[{x_n}\; = 2 + \left[ {1 + 2 + 4 + 7 + \ldots to{\rm{ }}\left( {n - 1} \right){\rm{ }}terms} \right]\]
Again let us consider that,
\[{S_2}\; = {\rm{ }}1 + 2 + 4 + 7 + \ldots {t_{n - 1}}\]------- (3)
\[{S_2}\; = 1 + 2 + 4 + 7 + \ldots {t_{n - 2}}\; + {\rm{ }}{t_{n - 1}}\]---- (4)
Now, we have to use the same method that is followed in solving equation (1) and equation (2)
Solve the same for equation (3) and equation (4), we get
\[0 = 1 + \left[ {1 + 2 + 3 + \ldots \left( {n - 2} \right)term} \right]-{\rm{ }}{t_{n - 1}}\]
Now, we have to determine the sum of n natural numbers\[ = \dfrac{{{\rm{ }}n\left( {n + 1} \right)}}{2}\]
Now, let us find
\[{t_{n - 1}}\; = {\rm{ }}\dfrac{{1{\rm{ }} + {\rm{ }}\left( {n - 2} \right)\left( {n - 1} \right)}}{2}\]
Now, we have to solve the numerator of the above equation, we get
\[ = {\rm{ }}\dfrac{{(2 + {n^2}-3n + 2)}}{2}\]
On solving further, we get
\[ = {\rm{ }}\dfrac{{({n^2} - 3n + 4)}}{2}\]
On calculating for \[{S_2}\] we have
\[{S_2}\; = {\sum _{n = 1}}^{n - 2\;}{t_{n - 1}}\; = \dfrac{1}{2}\sum {n^2}-\left( {\dfrac{3}{2}} \right)\sum n{\rm{ }} + {\rm{ }}2\sum 1\]
On substituting the formulas, we get
\[ = \dfrac{1}{2}\left[ {\dfrac{{\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6}} \right]-\left[ {\left( {\dfrac{3}{2}} \right)\dfrac{{n\left( {n - 1} \right)}}{2}} \right] + 2\left( {n - 1} \right)\]
On further simplification of the above expression, we get
\[ = {\rm{ }}\left( {n - 1} \right)\left[ {\dfrac{{(2{n^2} - n)}}{{12}}-\left( {\dfrac{{3n}}{4}} \right) + 2} \right]\]
Now, let’s simplify the numerator to make it less complicated
\[ = {\rm{ }}\left( {n - 1} \right)\dfrac{{(2{n^2}\;-{\rm{ }}n{\rm{ }}-{\rm{ }}9n{\rm{ }} + {\rm{ }}24)}}{{12}}\]
\[ = {\rm{ }}\dfrac{{\left( {n - 1} \right)({n^2}\;-{\rm{ }}5n{\rm{ }} + {\rm{ }}12)}}{6}\]
Now, we have to multiply \[\left( {n - 1} \right)\] with each terms inside the parentheses, we get
\[ = \dfrac{{{\rm{ }}({n^3}\;-{\rm{ }}6{n^2}\; + {\rm{ }}17n{\rm{ }}-{\rm{ }}12)}}{6}\]
Hence, we have
\[{X_n} = 2 + {S_2}\]
\[ = 2 + ({n^3}-6{n^2} + 17n-12)\]
We already knew that,
\[ = {\rm{ }}\dfrac{{({n^3}\;-{\rm{ }}6{n^2}\; + {\rm{ }}17n)}}{6}\]
Now, we have to substitute the value of \[n\] we get
\[{x_{20}} = \dfrac{{({{20}^3}-6 \times {{20}^2}\; + {\rm{ }}17 \times 20)}}{6}\]
On simplifying the expressions inside the parentheses, we obtain
\[ = {\rm{ }}\dfrac{{\left( {8000{\rm{ }}-{\rm{ }}2400{\rm{ }} + {\rm{ }}340} \right)}}{6}\]
On further simplification, we get
\[ = 990\]
Therefore, in the series 2 + 3 + 5 + 9 + 16 +......, the 20th term is \[{\rm{990}}\]

Option ‘C’ is correct

Note: Because there are three types of series, namely Arithmetic, Geometric, and Harmonic, students should not be confused about which series to apply in which type of inquiry, as there are numerous formulas associated with each of the series. Students are more prone to make mistakes if they do not comprehend the question's phrasing and calculate the sum of the first 20 terms of the series, resulting in an incorrect solution.