Question

Find the sum ‘\[\sum\limits_{k = 1}^n {k(k + 2)} \] for any integer ‘\[n \ge 1\]’.
A. \[\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{2}\]
B. \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
C. \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 7} \right)}}{6}\]
D. \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 9} \right)}}{6}\]

Answer 1

Hint: First simplify the given expression and then solve using the formulas of summation of series.

Formula Used:
Summation formula \[\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] and \[\sum n = \dfrac{{n(n + 1)}}{2}\]
Expansion of sum of \[n\] term of natural number is \[1 + 2 + 3 + 4 + 5 + .... + n\]
Expansion of sum of \[n\] term of squared form is \[{1^2} + {2^2} + {3^2} + {4^2} + {5^2} + .... + {n^2}\]
Summation formula \[\sum {\left( {n(x) + p(x)} \right)} = \sum {n(x)} + \sum {p(x)} \]

Complete step by step solution:
For any integer \[n \ge 1\], the sum \[\sum\limits_{k = 1}^n {k(k + 2)} \] is equal to:
\[\sum\limits_{k = 1}^n {k(k + 2)} \]
Simplifying the above expression and we get
\[ = \sum\limits_{k = 1}^n {\left( {{k^2} + 2k} \right)} \]
Using the summation formula \[\sum {\left( {n(x) + p(x)} \right)} = \sum {n(x)} + \sum {p(x)} \] and we get
\[ = \sum\limits_{k = 1}^n {{k^2}} + \sum\limits_{k = 1}^n {2k} \]
Taking constant part outside of the summation
\[ = \sum\limits_{k = 1}^n {{k^2}} + 2\sum\limits_{k = 1}^n k \]
Now, expanding the above summation and we get
\[ = {1^2} + {2^2} + {3^2}....... + 2\left( {1 + 2 + 3 + ......n} \right)\]
Using the summation formulas \[\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] and \[\sum n = \dfrac{{n(n + 1)}}{2}\], we get
\[ = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2 \times \dfrac{{n\left( {n + 1} \right)}}{2}\]
Taking common from the above expression and we get
\[ = n\left( {n + 1} \right)\left( {\dfrac{{\left( {2n + 1} \right)}}{6} + 2 \times \dfrac{1}{2}} \right)\]
\[ = n\left( {n + 1} \right)\left( {\dfrac{{2n + 1}}{6} + 1} \right)\]
\[ = n\left( {n + 1} \right)\left( {\dfrac{{2n + 1 + 6}}{6}} \right)\]
\[ = n\left( {n + 1} \right)\left( {\dfrac{{2n + 7}}{6}} \right)\]
\[ = \dfrac{{n\left( {n + 1} \right)\left( {2n + 7} \right)}}{6}\]

The correct answer is option C.

Note: A series of supplied numbers are added together to produce their sum or total in the summation process. It is typically necessary when a lot of data is provided and it tells you to add up all the values in a certain order. Summation is a crucial mathematical concept since it calculates numerous terms in a given sequence. The representation of huge numbers requires summation notation. To put it another way, summation notation allows us to express the addition of very big values in a sequence.