Question

If \[{S_1},{S_2},{S_3},....{S_n}\] be the sum of the n terms of n A.P.s whose first terms are 1,2,3,…..n respectively and common differences are 1,3,5,….. \[2n - 1\] respectively. Then \[\sum\limits_{r = 1}^n {{S_r}} \] is
A) \[\dfrac{{{n^3}\left( {n + 1} \right)}}{2}\]
B) \[\dfrac{{\left( {{n^2}\left( {{n^2} - 1} \right) + 2} \right)}}{2}\]
C) \[\dfrac{{n\left( {{n^3} + 1} \right)}}{2}\]
D) \[\dfrac{{{n^2}\left( {{n^2} + 1} \right)}}{2}\]

Answer 1

Hint: The question is complicated from the wordings but it is too easy to solve. We will use the same formula that we use to solve the sum of n terms of an A.P. rather here we will add the A.P. directly. So just be careful when we add the A.P.s in between we will need the formula for finding the sum of consecutive natural numbers and the sum of odd numbers.

Complete step by step answer:
Given that the sum are \[{S_1},{S_2},{S_3},....{S_n}\]
Now as we know that the formula for the sum of n terms of an A.P is given by,
  \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
We will apply this for the A.P. one by one,
\[\Rightarrow {S_1} = \dfrac{n}{2}\left[ {2 + \left( {n - 1} \right)1} \right]\]
\[\Rightarrow {S_2} = \dfrac{n}{2}\left[ {4 + \left( {n - 1} \right)3} \right]\]
\[\Rightarrow {S_3} = \dfrac{n}{2}\left[ {6 + \left( {n - 1} \right)5} \right]\]
This continues up to the nth sum of A.P.
\[\Rightarrow {S_n} = \dfrac{n}{2}\left[ {2n + \left( {n - 1} \right)\left( {2n - 1} \right)} \right]\]
Now we will add these sums to get the actual solution.
\[{S_1} + {S_2} + {S_3} + ...{S_n} = \dfrac{n}{2}\left[ {2 + \left( {n - 1} \right)1} \right] + \dfrac{n}{2}\left[ {4 + \left( {n - 1} \right)3} \right] + \dfrac{n}{2}\left[ {6 + \left( {n - 1} \right)5} \right] + .....\dfrac{n}{2}\left[ {2n + \left( {n - 1} \right)\left( {2n - 1} \right)} \right]\]
Now the LHS can be simply written is summation form,
\[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{n}{2}\left[ {2 + \left( {n - 1} \right)1} \right] + \dfrac{n}{2}\left[ {4 + \left( {n - 1} \right)3} \right] + \dfrac{n}{2}\left[ {6 + \left( {n - 1} \right)5} \right] + .....\dfrac{n}{2}\left[ {2n + \left( {n - 1} \right)\left( {2n - 1} \right)} \right]\]
Now we will separate the sum of first terms and the common differences as,
\[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{n}{2}\left( {2 + 4 + 6 + .....2n} \right) + \dfrac{n}{2}\left( {n - 1} \right)\left( {1 + 3 + 5 + ...\left( {2n - 1} \right)} \right)\]
In the above expression the first term gives the first terms and the later bracket gives the common differences.
Now from first bracket we will take 2 common and cancel it with the 2 outside,
\[\sum\limits_{r = 1}^n {{S_r}} = n\left( {1 + 2 + 3 + ....n} \right) + \dfrac{n}{2}\left( {n - 1} \right)\left( {1 + 3 + 5 + ...\left( {2n - 1} \right)} \right)\]
Now for the first bracket \[1 + 2 + 3 + ....n\] we will use the formula to find the sum of n natural numbers and for the second bracket \[1 + 3 + 5 + ...\left( {2n - 1} \right)\] we will use the formula to find the sum of odd numbers.
\[\sum\limits_{r = 1}^n {{S_r}} = n\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{n}{2}\left( {n - 1} \right){\left( {\dfrac{1}{2}\left( {2n - 1 + 1} \right)} \right)^2}\]
On solving the bracket from the second term,
\[\sum\limits_{r = 1}^n {{S_r}} = n\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{n}{2}\left( {n - 1} \right){\left( {\dfrac{1}{2}\left( {2n} \right)} \right)^2}\]
\[\Rightarrow \sum\limits_{r = 1}^n {{S_r}} = \dfrac{{{n^2}\left( {n + 1} \right)}}{2} + \dfrac{n}{2}\left( {n - 1} \right){\left( n \right)^2}\]
Taking \[\dfrac{{{n^2}}}{2}\] common,
\[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{{{n^2}}}{2}\left[ {\left( {n + 1} \right) + n\left( {n - 1} \right)} \right]\]
Multiplying n with the bracket,
\[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{{{n^2}}}{2}\left[ {\left( {n + 1} \right) + {n^2} - n} \right]\]
Cancelling the opposite terms,
\[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{{{n^2}}}{2}\left[ {1 + {n^2}} \right]\]
Therefore, the sum of the given A.P.s is \[\sum\limits_{r = 1}^n {{S_r}} = \dfrac{{{n^2}}}{2}\left[ {1 + {n^2}} \right]\]. Thus, option (D) is correct.

Note:
Note that, we have added the sum of all the A.P.s given but the first term s and common differences were given separately. Since coincidently they can be solved by the formula we use to find the sum of natural numbers and odd numbers. Thus was a bit long but not confusing.
Also note that, we just separated the formula to find the sum of n terms of an A.P. in two separate parts, nothing else!