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.