Question

How do you find the sum of $\sum {{{(k + 1)}^2}(k - 3)} $ where “k” is $\left[ {2,\;5} \right]?$

Answer 1

Hint: The question is asking for the sum of $\sum {{{(k + 1)}^2}(k - 3)} $ from $k = 2\;{\text{to}}\;5$, so this is not a big deal to find the sum as the interval of “k” is too small, simply put value of “k” as integers from $2\;{\text{to}}\;5$ in the given general term of the series and add them, you will get the required sum.

Complete step by step solution:
In order to find the sum of $\sum {{{(k + 1)}^2}(k - 3)} $ from $k = 2\;{\text{to}}\;5$, we will simply put all the values of “k” as integers from $2\;{\text{to}}\;5$ in the given general term of the series as follows
Firstly we will write the values of “k” in set builder form
So, $k = \left[ {2,\;5} \right] = \left\{ {2,\;3,\;4,\;5} \right\}$
And general term of the series: ${(k + 1)^2}(k - 3)$
Now, we will find terms by putting value of “k” from $2\;{\text{to}}\;5$ one by one
\[
  k = 2, \Rightarrow {(2 + 1)^2}(2 - 3) = {3^2} \times ( - 1) = - 9 \\
  k = 3, \Rightarrow {(3 + 1)^2}(3 - 3) = {4^2} \times 0 = 0 \\
  k = 4, \Rightarrow {(4 + 1)^2}(4 - 3) = {5^2} \times 1 = 25 \\
  k = 5, \Rightarrow {(5 + 1)^2}(5 - 3) = {6^2} \times 2 = 36 \times 2 = 72 \\
 \]
Now, adding them all we will get
\[\sum\limits_{k = 2}^{k = 5} {{{(k + 1)}^2}(k - 3)} = - 9 + 0 + 25 + 72 = 88\]

Therefore \[88\] is the required sum for the given series.

Note: Use this method only when the interval is very small, if you will use this process for an interval having length of more than ten then it will be lengthy for you to calculate the required sum, as you have to first calculate the individual terms.
But do not worry we have one more way for you, which is the standard one in which we use summation formulas to directly calculate the sum by putting lower and upper limits of the interval. In this method we first simplify the general term of the series and then use a summation formula for each term and eventually derive a summation formula for the series. Try this question from the summation method yourself.