Question

Find the $1 \cdot 4 \cdot 7 + 2 \cdot 5 \cdot 8 + 3 \cdot 6 \cdot 9 + ......... + {n^{th}}$ terms.

Answer 1

Start with finding the general term or ${n^{th}}$ term for the given series. Notice the pattern to write the expression for ${n^{th}}$ term in terms of ‘n’ only. After finding the general term, apply the sigma operator to represent the sum of the series. Now substitute the values $\sum n = \dfrac{{n\left( {n + 1} \right)}}{2},\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ and }}\sum {{n^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$ in the expression to get the expression for the sum of the series.

Complete step-by-step answer:
Here in this problem, we are given a series of the sum of the product of three numbers. We need to find the sum of the first ‘n’ number of terms of this series.
In the given series $1 \cdot 4 \cdot 7 + 2 \cdot 5 \cdot 8 + 3 \cdot 6 \cdot 9 + ......... + {n^{th}}$ term, we need to first find out the ${n^{th}}$ term
We can see the pattern in the series that each term is the product of three numbers as follows:
$\Rightarrow {\text{ First Term}} = 1 \cdot 4 \cdot 7 = \left( {0 + 1} \right) \cdot \left( {3 + 1} \right) \cdot \left( {6 + 1} \right)$
$\Rightarrow {\text{ Second Term}} = 2 \cdot 5 \cdot 8 = \left( {1 + 1} \right) \cdot \left( {4 + 1} \right) \cdot \left( {7 + 1} \right) = \left( 2 \right) \cdot \left( {3 + 2} \right) \cdot \left( {6 + 2} \right)$
$\Rightarrow {\text{ Third Term}} = 3 \cdot 6 \cdot 9 = \left( {2 + 1} \right) \cdot \left( {5 + 1} \right) \cdot \left( {8 + 1} \right) = \left( 3 \right) \cdot \left( {3 + 3} \right) \cdot \left( {6 + 3} \right)$
$\Rightarrow {\text{ Fourth Term}} = 4 \cdot 7 \cdot 10 = \left( {3 + 1} \right) \cdot \left( {6 + 1} \right) \cdot \left( {9 + 1} \right) = \left( 4 \right) \cdot \left( {3 + 4} \right) \cdot \left( {6 + 4} \right)$
Similarly using this pattern, we can find the general term of the given series as:
$\Rightarrow {\text{ }}{{\text{n}}^{th}}{\text{ term}} = \left( n \right) \cdot \left( {3 + n} \right) \cdot \left( {6 + n} \right)$
Therefore, now the given series can be represented as:
$\Rightarrow 1 \cdot 4 \cdot 7 + 2 \cdot 5 \cdot 8 + 3 \cdot 6 \cdot 9 + ......... + {n^{th}}{\text{ }}term = \sum {\left( n \right) \cdot \left( {3 + n} \right) \cdot \left( {6 + n} \right)}$
Here the symbol of summation represents the sum of each term with the value of ‘n’ all the natural numbers from $1{\text{ to }}n$ .
Now the summation expression can also be represented as
$\Rightarrow \sum {\left( n \right) \cdot \left( {3 + n} \right) \cdot \left( {6 + n} \right)} = \sum {\left( n \right) \cdot \left( {18 + 3n + 6n + {n^2}} \right)} = \sum {\left( {{n^3} + 9{n^2} + 18n} \right)}$
As we know that the summation or sigma is distributive on addition, therefore we can write the above expression as:
$\Rightarrow \sum {\left( {{n^3} + 9{n^2} + 18n} \right)} = \sum {\left( {{n^3}} \right)} + \sum {\left( {9{n^2}} \right)} + \sum {\left( {18n} \right)}$
Since the constant coefficients with a variable can be taken out of the sigma, we get:
$\Rightarrow \sum {\left( {{n^3} + 9{n^2} + 18n} \right)} = \sum {\left( {{n^3}} \right)} + \sum {\left( {9{n^2}} \right)} + \sum {\left( {18n} \right)} = \sum {\left( {{n^3}} \right)} + 9\sum {\left( {{n^2}} \right)} + 18\sum {\left( n \right)}$
So now we can easily calculate the sum of the series if we know the values for $\sum {\left( {{n^3}} \right)} ,\sum {\left( {{n^2}} \right)} \& \sum {\left( n \right)}$ .
According to the known values of summation of natural number series, we know that:
$\Rightarrow \sum n = \dfrac{{n\left( {n + 1} \right)}}{2},\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ and }}\sum {{n^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$
Let’s substitute these above values in the equation:
$\Rightarrow \sum {\left( {{n^3}} \right)} + 9\sum {\left( {{n^2}} \right)} + 18\sum {\left( n \right)} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + 9 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 18 \times \dfrac{{n\left( {n + 1} \right)}}{2}$
We can solve it further by taking common from the right side
$\Rightarrow \sum {\left( {{n^3}} \right)} + 9\sum {\left( {{n^2}} \right)} + 18\sum {\left( n \right)} = n\left( {n + 1} \right)\left( {\dfrac{{n\left( {n + 1} \right)}}{4} + 3 \times \dfrac{{\left( {2n + 1} \right)}}{2} + 9} \right)$
Now we can simplify the parenthesis on the right side of the equation:
$\Rightarrow \sum {\left( {{n^3}} \right)} + 9\sum {\left( {{n^2}} \right)} + 18\sum {\left( n \right)} = n\left( {n + 1} \right)\left( {\dfrac{{{n^2} + n + 12n + 6 + 36}}{4}} \right) = \dfrac{{n\left( {n + 1} \right)\left( {{n^2} + 13n + 42} \right)}}{4}$
If we use the middle term split method, then we can simplify the quadratic expression and write it as:
$\Rightarrow \sum {\left( {{n^3}} \right)} + 9\sum {\left( {{n^2}} \right)} + 18\sum {\left( n \right)} = \dfrac{{n\left( {n + 1} \right)\left( {{n^2} + 6n + 7n + 42} \right)}}{4} = \dfrac{{n\left( {n + 1} \right)\left( {n + 6} \right)\left( {n + 7} \right)}}{4}$
Therefore, we get the expression for the sum of the given series as:
$\Rightarrow 1 \cdot 4 \cdot 7 + 2 \cdot 5 \cdot 8 + 3 \cdot 6 \cdot 9 + ......... + {n^{th}}{\text{ }}term = \dfrac{{n\left( {n + 1} \right)\left( {n + 6} \right)\left( {n + 7} \right)}}{4}$

Note: In questions like this, where the progression is neither arithmetic nor geometric, finding the general term plays an important role. The use of a summation or sigma operator must be done carefully. After getting the sum of the series of ‘n’ terms as $\dfrac{{n\left( {n + 1} \right)\left( {n + 6} \right)\left( {n + 7} \right)}}{4}$, we can cross-check our conclusion by putting $n = 1$ , which should give the value of the first term, i.e. $1 \times 4 \times 7 = 28$ . For $n = 1$ ; $\dfrac{{n\left( {n + 1} \right)\left( {n + 6} \right)\left( {n + 7} \right)}}{4} = \dfrac{{1 \times 2 \times 7 \times 8}}{4} = 2 \times 7 \times 2 = 28$ . Thus our answer is correct.