Question

Find the sum of the series for the first n brackets $ (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + \_ $ is
A. $ {(n - 1)^3} + {n^3} $
B. $ {(n - 1)^3} + 8{n^2} $
C. $ \dfrac{{(n + 1)(n + 2)}}{6} $
D. $ \dfrac{{(n + 3)(n + 2)}}{{12}} $

Answer 1

Hint: The arithmetic series is the sum of all the terms of the arithmetic progression (AP). Where, “a” is the first term and “d” is the common difference among the series.
 $
  {S_n}{\text{ = a + (a + d) + (a + 2d) + (a + 3d) + }}......{\text{ + [a + (n - 1)d]}} \\
  {{\text{S}}_{n{\text{ }}}} = \dfrac{n}{2}[2a + (n - 1)d] \\
  {{\text{S}}_{n{\text{ }}}} = \dfrac{n}{2}[a + l] \\
  $
Where “a” is the first term and “l” is the last time.

Complete step-by-step answer:
Given series is - $ (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + \_ $
Just observe the above series and take out the very first term of each of the brackets.
 $ 1,2,5,10,.......{t_n} $
Now, take the sum of the above terms by using the summation of the series.
 $ {S_n} = 1 + 2 + 5 + 10 + ....... + {t_n}{\text{ }}.....{\text{ (i)}} $
The above equation can be re-written as by adding zero ahead of the addition of all other terms.
 $ {S_n} = 0 + 1 + 2 + 5 + 10 + ....... + {t_{n - 1}}{\text{ + }}{t_n}{\text{ }}.....{\text{ (ii)}} $
Subtract the equation (ii) from the equation (i)
 $ \Rightarrow 0 = 1 + 1 + 3 + 5 + ..... + (n - 1){\text{ terms - }}{{\text{t}}_n} $
When any term is moved from one side to the other side, the sign also changes from positive to negative and vice versa.
 $ \Rightarrow {t_n} = 1 + [1 + 3 + 5 + ..... + (n - 1){\text{ terms ]}} $
Let us assume that –
 $ \Rightarrow {t_n} = 1 + {S_n} $ ...... (A)
Now, observe the second part of the equation- here all the terms have common differences so apply formula for the summation of the arithmetic progression.
 $ {S_n} = \dfrac{n}{2}[2a + (n - 1)d] $
Place the values in the standard formula –
Where, $ a = 1{\text{ and d = 2 also, n = n - 1}} $
 $ {S_n} = \dfrac{{n - 1}}{2}[2(1) + (n - 1 - 1)2] $
Simplify the above equation –
 $
  {S_n} = \dfrac{{n - 1}}{2}[2 + (n - 2)2] \\
   \Rightarrow {S_n} = \dfrac{{n - 1}}{2}[2 + 2n - 4] \;
  $
Directly add or subtract the like terms-
 $ \Rightarrow {S_n} = \dfrac{{n - 1}}{2}[2n - 2] $
Take common factor in the above equation-
 $ \Rightarrow {S_n} = \dfrac{{n - 1}}{2} \times 2[n - 1] $
Same terms from the numerator and the denominator cancel each other.
 $
   \Rightarrow {S_n} = (n - 1)(n - 1) \\
   \Rightarrow {S_n} = {(n - 1)^2} \;
  $
Place the above value in the equation (A)
 $ {t_n} = 1 + {(n - 1)^2} $
Expand the above bracket-
 $ {t_n} = 1 + {n^2} - 2n + 1 $
Simplify the above equation –
 $ {t_n} = {n^2} - 2n + 2 $ this is the last term of the series.
Here, we want the total number of terms $ = (2n - 1) $
Also, $ d = 1, $ and $ {t_n} = a + (n - 1)d $
Now, the series becomes –
 $ {t_n},{\text{ }}{t_{n + 1}},{\text{ }}{{\text{t}}_{n + 2}}.......(2n - 1){\text{ }}terms $
Here $ a = {t_n},\;{\text{n = (2n - 1) }}\;{\text{d = 1}} $
Place the above values in the standard formula-
 $ \Rightarrow {t_n} + (2n - 1)(1) $
Place values by using the equation (A) in the above equation –
 $ \Rightarrow {n^2} - 2n + 2 + (2n - 1)(1) $
Simplify the above equation-
 $ \Rightarrow {n^2} - 2n + 2 + 2n - 1 $
Terms with same value and opposite sign cancel each other
 $ \Rightarrow {n^2} + 1 $
Now, summation of the terms with $ (2n - 1) $ terms-
Sum $ = \dfrac{n}{2}[a + l] $
Here, $ a = {t_n},\;l{\text{ = }}{{\text{n}}^2} + 1{\text{ and n = (2n - 1)}} $
Place in the above equation-
Sum $ = \dfrac{{2n - 1}}{2}[{n^2} - 2n + 2 + {n^2}] $
Simplify the above equation-
Sum $ = \dfrac{{2n - 1}}{2}[2{n^2} - 2n + 2] $
Take common factor –
Sum $ = \dfrac{{2n - 1}}{2} \times 2[{n^2} - n + 1] $
Same terms from the numerator and the denominator cancel each other.
Sum $ = (2n - 1)[{n^2} - n + 1] $
Simplify the above equation –
Sum
  $
   = 2n[{n^2} - n + 1] - 1[{n^2} - n + 1] \\
   = 2{n^3} - 2{n^2} + 2n - {n^2} + n - 1 \\
   = 2{n^3} - 3{n^2} + 3n - 1 \;
  $
The above equation can be written as-
By using the standard formula - $ {(a - b)^3} = {a^3} - {b^3} - 3ab(a - b) $
 $
  2{n^3} - 3{n^2} + 3n - 1 = {n^3} + \underline {{n^3} - 3{n^2} + 3n - 1} \\
  2{n^3} - 3{n^2} + 3n - 1 = {n^3} + \underline {{n^3} - {{(1)}^3} - 3n(n - 1)} \\
  2{n^3} - 3{n^2} + 3n - 1 = {n^3} + {(n - 1)^3} \\
  $
So, the correct answer is “Option A”.

Note: The above question can be solved by using trial and error method taking the given multiple options and place the values in it.
For Example: Take option (A) given as $ {(n - 1)^3} + {n^3} $
For $ n = 1 $
Given that the sum of the terms in the first bracket is $ = 1 $
Place the values-
 $ {(n - 1)^3} + {n^3} = {(1 - 1)^3} + {(1)^3} = 0 + 1 = 1 $
For $ n = 2 $
Given that the sum of the terms in the first bracket is $ = 2 $
Place the values-
 $ {(n - 1)^3} + {n^3} = {(2 - 1)^3} + {(2)^3} = 1 + 8 = 9 $
For $ n = 3 $
Given that the sum of the terms in the first bracket is $ = 2 $
Place the values-
 $ {(n - 1)^3} + {n^3} = {(3 - 1)^3} + {(3)^3} = 8 + 27 = 35 $
Hence, the taken option (A) is the correct answer.