Question Video: Finding a Formula for the Sum from 1 to 𝑛 of 𝑟 | Nagwa Question Video: Finding a Formula for the Sum from 1 to 𝑛 of 𝑟 | Nagwa

Question Video: Finding a Formula for the Sum from 1 to 𝑛 of π‘Ÿ Mathematics • Second Year of Secondary School

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Complete the following: βˆ‘_(π‘Ÿ = 1) ^(𝑛) π‘Ÿ = οΌΏ.

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Video Transcript

Complete the following. The sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to blank.

So what we’re trying to find is an algebraic expression for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ. If we were to write this out longhand, it is one plus two plus three and so on up to 𝑛 minus one plus 𝑛. Now, in order to find the formula we’re looking for, there is a trick which is that we can write this sum out in two ways. We can switch the order of the terms around and write this equivalently as 𝑛 plus 𝑛 minus one plus 𝑛 minus two all the way down to plus two plus one.

Now, what we might notice is that by swapping the order of the terms around, the sum of the terms in each position is always the same. Each time, the sum is equal to 𝑛 plus one. So if we were to add to these two expressions for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ together, we would have two lots of the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 plus one plus 𝑛 plus one plus 𝑛 plus one and so on. In fact, we’d be adding 𝑛 lots of 𝑛 plus one, which can be written algebraically as 𝑛 multiplied by 𝑛 plus one.

We want to find an expression for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ, whereas at the moment we have two lots of this. We can therefore divide both sides of this equation by two to give the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 multiplied by 𝑛 plus one over two. And this is our answer. The sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ, which is the sum of the integers from one to 𝑛, is 𝑛 multiplied by 𝑛 plus one over two. We should commit this result to memory but we should also remember how to derive it.

Let’s also look at an alternative approach we could take to answering this question. We’re going to consider the binomial expansion of π‘Ÿ minus one all squared. And the reason for doing this will become clear as we work through the method. π‘Ÿ minus one squared is equal to π‘Ÿ squared minus two π‘Ÿ plus one. Rearranging, we have that two π‘Ÿ minus one is equal to π‘Ÿ squared minus π‘Ÿ minus one squared. And as these two expressions are equal, the sum from π‘Ÿ equals one to 𝑛 of the series with each expression as its general term will also be equal.

Let’s consider the sum on the right-hand side first. If we were to write this sum out longhand, we would have one squared minus zero squared plus two squared minus one squared plus three squared minus two squared and so on out to 𝑛 squared minus 𝑛 minus one squared. What we’ll notice though is that several of the terms will cancel out. First, we’re adding one squared, and then we’re subtracting one squared. We’re adding two squared and then subtracting two squared and so on. In fact, if we continue in this way, the only terms that are left are the negative zero squared when π‘Ÿ is equal to one and 𝑛 squared when π‘Ÿ is equal to 𝑛. But of course, zero squared is just equal to zero. So in fact, the entire sum simplifies to 𝑛 squared.

Next, let’s consider the sum on the left-hand side. Using the summation linearity property, the sum from π‘Ÿ equals one to 𝑛 of two π‘Ÿ minus one is equal to two multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ minus the sum from π‘Ÿ equals one to 𝑛 of one. In this second term, we are simply adding the value one together 𝑛 times. So the value of the second sum is one multiplied by 𝑛, which is equal to 𝑛. So equating the expressions for the sums on each side of the equation, we have that two multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ minus 𝑛 is equal to 𝑛 squared.

We want to rearrange this equation to give an expression for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ. We begin by adding 𝑛 to each side to give two multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 squared plus 𝑛. This can be factored by 𝑛 to give 𝑛 multiplied by 𝑛 plus one. The final step is to divide both sides of the equation by two, giving the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 multiplied by 𝑛 plus one all over two, which is the same as the formula we found using our previous method.

So by first using a neat trick of writing the series out in two different orders and then by considering the binomial expansion of π‘Ÿ minus one squared, we found that the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 multiplied by 𝑛 plus one over two.

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