Question

How do you use the properties of summation to evaluate the sum of $\sum{{{i}^{3}}-2i}$ from $i=1$ to $15$ ?

Answer 1

We first break down the entire summation into two summations $\sum\limits_{i=1}^{15}{{{i}^{3}}}-\sum\limits_{i=1}^{15}{2i}$ using the property. Then, the first summation can be evaluated using the formula for the sum of cubes of natural numbers and the second summation can be solved using the sum of natural numbers formula.

Complete step by step solution:
The symbol $\sum{{}}$ means summation. Generally, the infinite or finite series summation are denoted by the symbol $\sum{{}}$ . This summation contains a single variable which attains values from an initial value to a final value. For example, $\sum\limits_{i=1}^{10}{i}$ means the summation of all-natural numbers from $1$ to $10$ .
In this problem, the given summation that we need to evaluate is $\sum{{{i}^{3}}-2i}$ from $i=1$ to $15$ . This can be rewritten as,
$\Rightarrow \sum\limits_{i=1}^{15}{{{i}^{3}}-2i}$
Now, there is a property of summation operator. This property states that if the summation term is an addition or subtraction of multiple terms, then the entire summation can be broken down into separate summations. For example, $\sum{ai+b}=\sum{ai}+\sum{b}$ . Applying this property to the given summation, the summation thus becomes,
$\Rightarrow \sum\limits_{i=1}^{15}{{{i}^{3}}}-\sum\limits_{i=1}^{15}{2i}$
The first summation of the expression is simply the summation of the cubes of natural numbers. Now, there is a formula for the summation of consecutive natural number from $1$ which is ${{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$ , where n is the number of natural numbers starting from $1$ taken into account. So, the first summation can be written as,
$\Rightarrow {{\left( \dfrac{15\left( 15+1 \right)}{2} \right)}^{2}}-\sum\limits_{i=1}^{15}{2i}$
That can be simplified to,
$\Rightarrow 14400-\sum\limits_{i=1}^{15}{2i}$
There is another property of summations which is $\sum{ai}=a\sum{i}$ where a is a constant. So, the summation becomes,
$\Rightarrow 14400-2\sum\limits_{i=1}^{15}{i}$
The second summation is the summation of first $15$ natural numbers which has the formula $\dfrac{n\left( n+1 \right)}{2}$ where n is $15$ . The summation becomes,
$\begin{aligned} & \Rightarrow 14400-2\times \dfrac{15\left( 15+1 \right)}{2} \\\ & \Rightarrow 14400-240 \\\ & \Rightarrow 14160 \\\ \end{aligned}$

Therefore, we can conclude that the given summation evaluates to $14160$

Note: The summation problems require patience. We should apply the properties of summation correctly and wisely in order to get to our final answer quickly. Also, we need to remember some basic summation formulae like the sum of n natural numbers, sum of their squares and so on, which may lead us to the final answer faster.