Question

The sum of $2 + 12 + 36 + + 80 + 150........ + {\text{n terms = }}$
$\left( 1 \right)\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{24}}$
$\left( 2 \right)\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}$
$\left( 3 \right)\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 5} \right)}}{{24}}$
$\left( 4 \right)\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{{12}}$

Answer 1

We have to find the sum of $n$ terms of the given series $2 + 12 + 36 + 80 + 150...... + n$ . The easy approach to solve this type of question should be analyzing the given series in the form of Arithmetic progression (A.P.) or Geometric progression (G.P.) by finding out some kind of pattern for the given series. But the given series is neither in A.P. nor in G.P. So, what should be the approach to solve this kind of problem? The complete step by step method is given below.

Complete step by step answer:
If the given sequence is neither an A.P. nor a G.P. ;
$\left( 1 \right)$ The first step should be to find the difference between the successive terms (first difference , second difference , third difference and so on..) and check whether the differences are in A.P. or not .
If the first successive difference is in the form of A.P. then, general term can be given as; ${T_n} = a{n^2} + bn + c$ (i.e. ${t_n}$ is a quadratic polynomial with decreasing powers of $n{\text{ with constants }}a,b,c$ ) .
$\left( 2 \right)$ If the first successive difference does not give us an A.P. and we have to find out the second successive difference and third successive difference. If the second and third successive differences are in A.P. then the general term can be given by ${T_n} = a{n^3} + b{n^2} + cn + d$ (i.e. a cubic polynomial or polynomial of degree $3$ ) .
Now, let us proceed with the above steps to solve this question;
The given series is $2 + 12 + 36 + 80 + 150...... + n$;
The first successive difference can be calculated as;
$\Rightarrow {\Delta _1} = 12 - 2,36 - 12,80 - 36,150 - 80$
$\Rightarrow {\Delta _1} = 10,24,44,70$
Check whether the first successive difference is in A.P. or not; the answer is no here;
Therefore, the second successive difference can be calculated as;
$\Rightarrow {\Delta _2} = 24 - 10,44 - 24,70 - 44$
$\Rightarrow {\Delta _2} = 14,20,26$
We can notice a pattern here that the common difference is $6$ .
Therefore, the above series is in an A.P. and it’s general term can be given by;
$\Rightarrow {T_n} = a{n^3} + b{n^2} + cn + d{\text{ }}......\left( 1 \right)$
Put $n = 1$ in above equation, we get;
$\Rightarrow {T_1} = 2 = a + b + c + d{\text{ }}......\left( 2 \right)$
Put $n = 2$ in above equation $\left( 1 \right)$ , we get;
$\Rightarrow {T_2} = 12 = 8a + 4b + 2c + d{\text{ }}......\left( 3 \right)$
Put $n = 3$ in above equation $\left( 1 \right)$ , we get;
$\Rightarrow {T_3} = 36 = 27a + 9b + 3c + d{\text{ }}......\left( 4 \right)$
Put $n = 4$ in above equation $\left( 1 \right)$ , we get;
$\Rightarrow {T_4} = 80 = 64a + 16b + 4c + d{\text{ }}.....\left( 5 \right)$
Now, subtracting equations $\left( 5 \right){\text{ and }}\left( 4 \right),{\text{ }}\left( 4 \right){\text{ and }}\left( 3 \right),{\text{ }}\left( 3 \right){\text{ and }}\left( 2 \right)$ , we get;
$\Rightarrow 7a + 3b + c = 10{\text{ }}......\left( 6 \right)$
$\Rightarrow 19a + 5b + c = 24{\text{ }}......\left( 7 \right)$
$\Rightarrow 37a + 7b + c = 44{\text{ }}......\left( 8 \right)$
Again, subtracting equations $\left( 8 \right){\text{ and }}\left( 7 \right),{\text{ }}\left( 7 \right){\text{ and }}\left( 6 \right)$ , we get;
$\Rightarrow 12a + 2b = 14{\text{ }}......\left( 9 \right)$
$\Rightarrow 18a + 2b = 20{\text{ }}......\left( {10} \right)$
Now, solving equation $\left( 9 \right){\text{ and }}\left( {10} \right)$ to get the values of $a{\text{ and }}b$ ;
Subtracting equation $\left( {10} \right){\text{ from }}\left( 9 \right)$ ;
$\Rightarrow 18a - 12a = 20 - 14$
So, $a = 1$
Now put the value of $a$ in equation $\left( 9 \right){\text{ to get the value of }}b$ ;
$\Rightarrow 12 + 2b = 14$
Therefore, $b = 1$
Now put the value of $a{\text{ and }}b{\text{ in equation }}\left( 6 \right){\text{ to get the value of }}c$;
$\Rightarrow 10 + c = 10$
So, $c = 0$
Now, put the value of $a,b,c{\text{ in equation }}\left( 2 \right){\text{ to get the value of }}d$ ;
$\Rightarrow 2 + d = 2$
$\therefore d = 0$
Now put the value of $a,b,c,d{\text{ in equation }}\left( 1 \right),{\text{to get the general term }}{t_n};$
$\Rightarrow {T_n} = {n^3} + {n^2}$
Since, we have to find the sum of $n - terms;$
$\Rightarrow {S_n} = \sum {{T_n}} = \sum {{n^3}} + \sum {{n^2}}$
We know that, the formula for the addition of squares of natural numbers is ;
\Rightarrow \sum {{n^2}} = \dfrac{{\left\\{ {n\left( {n + 1} \right)\left( {2n + 1} \right)} \right\\}}}{6}
And we know that, the formula for the addition of sum of cubes of natural numbers is given by;
\Rightarrow \sum {{n^3}} = {\left\\{ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right\\}^2}
Now, put these values in the formula of sum of $n - terms$;
\Rightarrow {S_n} = {\left\\{ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right\\}^2} + \dfrac{{\left\\{ {n\left( {n + 1} \right)\left( {2n + 1} \right)} \right\\}}}{6}
Simplifying the above expression;
\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left\\{ {\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{{2n + 1}}{3}} \right\\}
Further simplifying;
\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left\\{ {\dfrac{{3{n^2} + 3n + 4n + 2}}{6}} \right\\}
\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left\\{ {\dfrac{{3{n^2} + 7n + 2}}{6}} \right\\}
Therefore, the final expression for the given series for sum of $n - terms$ is;
$\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {3{n^2} + 7n + 2} \right)}}{{12}}$
To match the options given to us; we can further solve the above expression;
$\Rightarrow {S_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}$

So, the correct answer is “Option 2”.

Note: The above process is generalized but very lengthy and it should be only used in cases where we can not find any pattern related to the given series. However, for this question we do notice some kind of pattern for the given series; $2 + 12 + 36 + + 80 + 150........ + {\text{n terms }}$. Here $2 = {1^2} + {1^3}$ ;$12 = {2^2} + {2^3}$ ; $36 = {3^2} + {3^3}$ ; $80 = {4^2} + {4^3}$ and $150 = {5^2} + {5^3}$ . Therefore, the ${n^{th}}$ term can be given by; ${S_n} = \sum\limits_{n = 1}^n {\left( {{n^2} + {n^3}} \right)} = \sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n {{n^3}}$ and then we can follow the same approach as above to get to the final answer.