Question Video: Using the Sum of a Constant Series to Determine the Number of Terms in the Series Mathematics

Given that βˆ‘_(π‘Ÿ = 1)^(𝑛) (π‘Ÿ = 3), find the value of 𝑛.

03:16

Video Transcript

Given that the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ equals three, find the value of 𝑛.

Let’s begin by recapping what this notation means. In general, the sum from π‘Ÿ equals one to 𝑛 of π‘Ž sub π‘Ÿ denotes the sum of the series with general term π‘Ž sub π‘Ÿ, which will be a function of π‘Ÿ. If the starting index is one and the finishing index is 𝑛, then we’re summing π‘Ž sub one, π‘Ž sub two, π‘Ž sub three, and so on all the way up to π‘Ž sub 𝑛. In this question, π‘Ž sub π‘Ÿ is just equal to π‘Ÿ, so we’re summing the index itself.

The sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is just one plus two plus three all the way up to adding 𝑛. Now we’re told in fact that this sum is equal to three. So we could answer this question by trial and error by adding the integers starting from one until we obtain three. If 𝑛 were equal to one, then we would have the sum from π‘Ÿ equals one to one of π‘Ÿ, which is simply equal to one. If 𝑛 is equal to two, then we have the sum from π‘Ÿ equals one to two of π‘Ÿ, which is one plus two. That’s of course equal to three, which is the sum we were looking for. And so we can stop.

The value of 𝑛 such that the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to three is two.

An alternative way to answer this question would be to recall a standard result, which is that the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 multiplied by 𝑛 plus one over two. We require this sum to be equal to three. And so we can form an equation in 𝑛. 𝑛 multiplied by 𝑛 plus one over two is equal to three. To solve for 𝑛, we’ll begin by multiplying both sides of the equation by two, giving 𝑛 multiplied by 𝑛 plus one is equal to six. We then distribute the parentheses on the left-hand side, giving 𝑛 squared plus 𝑛 is equal to six and then subtract six from each side of the equation to give 𝑛 squared plus 𝑛 minus six is equal to zero.

We have a quadratic equation in 𝑛 which can be solved by factoring. The factored form of this quadratic equation is 𝑛 plus three multiplied by 𝑛 minus two is equal to zero. It follows then that either 𝑛 plus three is equal to zero or 𝑛 minus two is equal to zero. To solve the first of these equations, we subtract three from each side, giving 𝑛 equals negative three. And to solve the second, we add two to each side, giving 𝑛 equals two. These are both valid solutions to this quadratic equation, but we recall that 𝑛 represents an integer greater than or equal to one. And so negative three is not a possible value for 𝑛 in this context. We proceed with the value 𝑛 equals two.

So using trial and error and using the standard result for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ, we found that if the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to three, then the value of 𝑛 is two.

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