Question

The ${n^{th}}$ term of the series $2 + 4 + 7 + 11 + ....$ will be
$A)\dfrac{{{n^2} + n + 1}}{2}$
$B){n^2} + n + 2$
$C)\dfrac{{{n^2} + n + 2}}{2}$
$D)\dfrac{{{n^2} + 2n + 2}}{2}$

Answer 1

First from the problem given that, the series is $2 + 4 + 7 + 11 + ....$and we need to find its generalized form.
The given series is not an arithmetic progression or geometric progression; we cannot apply the AP and GP formula for this problem, because there is no common difference or common ratio is fixed.
The only hint they gave is four numbers in the form of addition and we will solve them accordingly.
Formula used:
The general finite ${n^{th}}$ term of the series is $1 + 2 + 3 + 4... + n = \dfrac{{n(n + 1)}}{2}$

Complete step-by-step solution:
Since the given series is $2 + 4 + 7 + 11 + ....$and we will try to convert that series into $1 + 2 + 3 + 4... + n = \dfrac{{n(n + 1)}}{2}$, because which is the only hint for the given series to further solve.
Let us fix ${S_n} = 2 + 4 + 7 + 11 + ....$(n terms) (sum of the at most term is n) and if the sum of the at most term is $n - 1$then we have ${S_{n - 1}} = 2 + 4 + 7 + 11 + ....$($n - 1$ terms) (the last term is $n - 1$)
Now by our assumption, we have two series which are ${S_n} = 2 + 4 + 7 + 11 + ....$ n terms and ${S_{n - 1}} = 2 + 4 + 7 + 11 + ....$ $n - 1$ terms
Now we are going to subtract these two series with the condition keeping the first term of the first series is taken out common (because if we subtract in the general method then all the values are canceling out then we may not get any conclusion)
Thus, we get $S = {S_n} - {S_{n - 1}} = 2 + (4 - 2) + (7 - 4) + (11 - 7) + ....$ (keeping the first term of the first series and subtracted with the second terms)
Hence, we get $S = {S_n} - {S_{n - 1}} = 2 + 2 + 3 + 4 + + ...n$ (where the last term is $n + [(n - 1) - (n - 1)]$because we kept the first term and canceled with the $n - 1$ term with the second terms)
Rewrite the series as $S = {S_n} - {S_{n - 1}} = 1 + (1 + 2 + 3 + 4 + ... + n)$
Hence, we know that $1 + 2 + 3 + 4... + n = \dfrac{{n(n + 1)}}{2}$ then we get $S = {S_n} - {S_{n - 1}} = 1 + (\dfrac{{n(n + 1)}}{2})$
Solving this we get $S = {S_n} - {S_{n - 1}} = 1 + (\dfrac{{n(n + 1)}}{2}) \Rightarrow 1 + \dfrac{{{n^2} + n}}{2} = \dfrac{{{n^2} + n + 2}}{2}$
Therefore option $C)\dfrac{{{n^2} + n + 2}}{2}$ is correct.

Note: The concept of AM and GM is useful when we have a common difference or common ratio, which is an arithmetic progression that can be given by $a,(a + d),(a + 2d),(a + 3d),...$ where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$where $a$ is the first term and $r$ is a common ratio.
Hence the given question is in the form of geometric progression.