Question

How do you use the properties of summation to evaluate the sum of $\sum {{{\left( {i - 1} \right)}^2}}$ from $i = 1$ to $20$?

Answer 1

We will first use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ to simplify ${\left( {i - 1} \right)^2}$ to ${i^2} - 2i + 1$. Then we break the entire summation i.e., $\sum {\left( {{i^2} - 2i + 1} \right)}$ into three summations i.e., $\sum {{i^2}} - \sum {2i} + \sum 1$. Then we will use the standard results to evaluate each term and then we will simplify it to find the result.

Complete step-by-step answer:
Generally, the infinite or finite series summation are denoted by the symbol $\Sigma$. This summation contains a single variable which attains values from an initial value to a final value.
In this problem, we have to find the sum of $\sum {{{\left( {i - 1} \right)}^2}}$ from $i = 1$ to $20$. This can be rewritten as $\sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}}$.
Using the identity, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we get
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = \sum\limits_{i = 1}^{20} {{i^2} - 2i + 1}$
Now, there is a property of the summation operator that states that if the summation term is in addition or subtraction of multiple terms, then the entire summation can be broken down into separate summations. Using this we can write,
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = \sum\limits_{i = 1}^{20} {{i^2}} - \sum\limits_{i = 1}^{20} {2i} + \sum\limits_{i = 1}^{20} 1$
Also, using the property of summation i.e., $\sum {ai = a\sum i }$ where $a$ is constant. We can write,
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = \sum\limits_{i = 1}^{20} {{i^2}} - 2\sum\limits_{i = 1}^{20} i + \sum\limits_{i = 1}^{20} 1$
Some standard results which we will use are as follows:
For summation of squares of natural numbers, we have
$\Rightarrow \sum\limits_{r = 1}^n {{r^2}} = \dfrac{1}{6}n\left( {n + 1} \right)\left( {2n + 1} \right)$
For summation of consecutive natural numbers, we have
$\Rightarrow \sum\limits_{r = 1}^n r = \dfrac{1}{2}n\left( {n + 1} \right)$
For summation of $1$, we have
$\Rightarrow \sum\limits_{r = 1}^n 1 = n$
Using this, we get
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = \dfrac{1}{6} \times 20 \times \left( {20 + 1} \right)\left( {2 \times 20 + 1} \right) - 2\left[ {\dfrac{1}{2} \times 20 \times \left( {20 + 1} \right)} \right] + 20$
On simplifying, we get
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = \dfrac{1}{6} \times 20 \times 21 \times 41 - 20 \times 21 + 20$
On calculation, we get
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = 2870 - 420 + 20$
On simplifying, we get
$\Rightarrow \sum\limits_{i = 1}^{20} {{{\left( {i - 1} \right)}^2}} = 2470$
Therefore, on evaluating $\sum {{{\left( {i - 1} \right)}^2}}$ from $i = 1$ to $20$, we get $2470$.

Note: The symbol $\Sigma$ means summation. The summation is defined as $\sum\limits_{k = 1}^n {{a_k}} = {a_1} + {a_2} + {a_3} + ... + {a_n}$, which is the main property of summation. We obtain a sequence and then by adding the sequence we obtain the value. There are also some standard results which we use to find the summation such as for summation of consecutive natural numbers, summation of squares of natural numbers, etc.