Question

The sum of the series \[3 \cdot 6\, + \,4 \cdot 7\, + 5 \cdot 8\, + ....\] upto \[\left( {n - 2} \right)\]terms
A. \[{n^3} + {n^2} + n + 2\]
B. \[\left( {\dfrac{1}{6}} \right)\left( {2{n^3} + 12{n^2} + 10n - 84} \right)\]
C. \[{n^3} + {n^2} + n\]
D. none of these

Answer 1

Hint: In this question, first we consider the given series as two factors in which both are in AP. Now we have to find the nth term of both the factors and then calculate the sum of the series.

Formula used:
1. \[{a_n} = a + \left( {n - 1} \right)d\]
2. \[\sum {{n^2} = \dfrac{{n\,\left( {n - 1} \right)\,\,\left( {2n + 1} \right)}}{6}} \]
3. \[\sum {n\, = \dfrac{{n\,\left( {n + 1} \right)}}{2}} \]

Complete step-by-step solution:
Given series is \[3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8.....\]
It has two factors the first factor is \[3,4,5...\] which has a common difference \[d = 1\]
Now we find the \[{n_{th}}\] term by the formula \[{a_n} = a + \left( {n - 1} \right)d\]:
\[
  {a_n} = 3 + \left( {n - 1} \right) \cdot 1 \\
   = 3 + \left( {n - 1} \right) \\
   = n + 2 \\
 \]
Now the second factor is \[6,7,8,...\] which has a common difference \[d = 1\]
Now we find the \[{n_{th}}\] term by the formula \[{a_n} = a + \left( {n - 1} \right)d\]:
\[
  {a_n} = 6 + \left( {n - 1} \right) \cdot 1 \\
   = 6 + \left( {n - 1} \right) \\
   = n + 5 \\
 \]
Thus, the \[{n_{th}}\]term of the given series is \[{a_n} = \left( {n + 2} \right)\left( {n + 5} \right)\]
Therefore, the sum of the \[{n_{th}}\] term of the given series is \[{S_n} = \sum\limits_{n = 1}^{n - 2} {\left( {n + 2} \right)\,\left( {n + 5} \right)} \]
By multiplying \[\left( {n + 2} \right)\] by \[\left( {n + 5} \right)\] we obtain:
\[
  {S_n} = \sum\limits_{n = 1}^{n - 2} {\,n\,} \left( {n + 5} \right) + \,2\left( {n + 5} \right) \\
   = \sum\limits_{n = 1}^{n - 2} {\,{n^2} + 5n\, + 2n\, + 10} \\
   = \sum\limits_{n = 1}^{n - 2} {{n^2} + 7n + 10} \\
 \]
By splitting the terms, we get
\[
  {S_n} = \sum\limits_{n = 1}^{n - 2} {{n^2} + } \,\sum\limits_{n = 1}^{n - 2} {\,7n\, + \sum\limits_{n = 1}^{n - 2} {\,10} } \\
   = \sum\limits_{n = 1}^{n - 2} {{n^2} + } \,7\,\,\sum\limits_{n = 1}^{n - 2} {\,n\, + \sum\limits_{n = 1}^{n - 2} {\,10} } \,\,...\left( 1 \right) \\
 \]
We know that sum of squares of first n natural numbers is given by \[\sum {{n^2} = \dfrac{{n\,\left( {n + 1} \right)\,\,\left( {2n + 1} \right)}}{6}} \] and the sum of the first n natural number is given by \[\sum {n\, = \dfrac{{n\,\left( {n + 1} \right)}}{2}} \]
So, \[{S_n} = \dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)\,\,\left( {2\left( {n - 2} \right) + 1} \right)}}{6} + 7\dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)}}{2} + 10\left( {n - 2} \right)\]
Now by simplifying the above equation, we get
\[
  {S_n} = \dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)\,\,\left( {2n - 4 + 1} \right)}}{6} + 7\left( {\dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)}}{2}} \right) + 10\left( {n - 2} \right) \\
   = \dfrac{{\left( {n - 2} \right)\left( {n - 1} \right)\left( {2n - 3} \right)}}{6} + 7\left( {\dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)}}{2}} \right) + 10\left( {n - 2} \right) \\
   = \dfrac{{\left( {n - 2} \right)\left( {2{n^2} - 3n - 2n + 3} \right)}}{6} + 7\left( {\dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)}}{2}} \right) + 10\left( {n - 2} \right) \\
   = \dfrac{{\left( {n - 2} \right)\left( {2{n^2} - 5n + 3} \right)}}{6} + 7\left( {\dfrac{{\left( {n - 2} \right)\,\left( {n - 1} \right)}}{2}} \right) + 10\left( {n - 2} \right) \\
 \]
Further solving we get,
\[
  {S_n} = \left( {n - 2} \right)\left[ {\dfrac{{2{n^2} - 5n + 3}}{6} + \dfrac{{7\left( {n - 1} \right)}}{2} + 10} \right] \\
   = \left( {n - 2} \right)\left[ {\dfrac{{2{n^2} - 5n + 3 + 21n - 21 + 60}}{6}} \right] \\
   = \left( {n - 2} \right)\left[ {\dfrac{{2{n^2} + 16n + 42}}{6}} \right] \\
   = \left( {\dfrac{1}{6}} \right)\left( {2{n^3} + 16{n^2} + 42n - 4{n^2} - 32n - 84} \right) \\
 \]
Therefore, the sum of series is \[\left( {\dfrac{1}{6}} \right)\left( {2{n^3} + 12{n^2} + 10n - 84} \right)\]
Hence, option(B) is correct

Note: Because there are n terms, students must take the summation when calculating the sum. There are 'n-2' in the terms, and the summation must take its limit from 1 to n-2. You should also be able to calculate the sum of the first n natural numbers and the sum of the squares of the first n natural numbers.