### Video Transcript

Given that the sum of 𝑟 from 𝑟
equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one divided by two and the sum
of 𝑟 squared from 𝑟 equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one
multiplied by two 𝑛 plus one all divided by six, use the properties of summation
notation sigma to find the sum of seven 𝑟 squared minus seven 𝑟 minus 21 from 𝑟
equals one to four.

We will begin by rewriting our
expression using the linear property of summation. This gives us seven multiplied by
the sum of 𝑟 squared from 𝑟 equals one to four minus seven multiplied by the sum
of 𝑟 from 𝑟 equals one to four minus 21 multiplied by the sum of one from 𝑟
equals one to four. We are given expressions in terms
of 𝑛 for the sum of 𝑟 and the sum of 𝑟 squared. We also recall that the sum of one
from 𝑟 equals one to 𝑛 is equal to 𝑛.

In this question, the value of 𝑛
is four. The first term in our expression
becomes seven multiplied by four multiplied by five multiplied by nine all divided
by six. This can be simplified as shown,
giving us seven multiplied by 30. The second term is equal to seven
multiplied by four multiplied by five divided by two. This is simply seven multiplied by
10. As the third term becomes 21
multiplied by four, we are left with seven multiplied by 30 minus seven multiplied
by 10 minus 21 multiplied by four. This is equal to 56. The sum of seven 𝑟 squared minus
seven 𝑟 minus 21 from 𝑟 equals one to four is 56.

Alternatively, we could’ve
substituted the integers from one to four into our expression and then found the sum
of these four values. When 𝑟 equals one, seven 𝑟
squared minus seven 𝑟 minus 21 is equal to negative 21. When 𝑟 equals two, the expression
equals negative seven. When 𝑟 equals three, we get an
answer of 21. And when 𝑟 equals four, the
expression equals 63. As these four numbers sum to give
us 56, this confirms our answer is correct.