Question Video: Deriving the Formula for the Sum from 1 to 𝑛 of 𝑟² | Nagwa Question Video: Deriving the Formula for the Sum from 1 to 𝑛 of 𝑟² | Nagwa

Question Video: Deriving the Formula for the Sum from 1 to 𝑛 of π‘ŸΒ² Mathematics • Second Year of Secondary School

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Complete the following: βˆ‘_(π‘Ÿ = 1)^(𝑛) π‘ŸΒ² = οΌΏ.

07:18

Video Transcript

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

In this question, we’re asked to find an algebraic expression for the sum of the series with general term π‘Ÿ squared from π‘Ÿ equals one to π‘Ÿ equals 𝑛. This will be an algebraic expression in terms of 𝑛. Now the answer to this is actually a standard result that we should commit to memory, but here we’ll work through the process of deriving this result. What we’re going to do may not be intuitively obvious straightaway, but it will become clear why we’ve taken this approach.

We’re going to consider the binomial expansion of π‘Ÿ minus one cubed. Evaluating this longhand or by applying the binomial theorem, we find that π‘Ÿ minus one cubed is equal to π‘Ÿ cubed minus three π‘Ÿ squared plus three π‘Ÿ minus one. Subtracting π‘Ÿ minus one cubed from each side, we have zero is equal to π‘Ÿ cubed minus π‘Ÿ minus one cubed minus three π‘Ÿ squared plus three π‘Ÿ minus one. And then adding three π‘Ÿ squared, subtracting three π‘Ÿ, and adding one to each side of the equation, we have three π‘Ÿ squared minus three π‘Ÿ plus one is equal to π‘Ÿ cubed minus π‘Ÿ minus one cubed.

Now as these expressions are equivalent, their sums from π‘Ÿ equals one to 𝑛 will also be equivalent. We’re going to consider them separately, starting with the expression on the right-hand side, the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ cubed minus π‘Ÿ minus one cubed. Writing this sum out longhand for π‘Ÿ equals one up to π‘Ÿ equals 𝑛, we have one cubed minus zero cubed plus two cubed minus one cubed plus three cubed minus two cubed all the way up to 𝑛 cubed minus 𝑛 minus one cubed.

What we notice is that several terms in this expression will cancel out. We have one cubed and then later we are subtracting one cubed. We are adding two cubed, and later we are subtracting two cubed. We continue in this way and we find that the only terms that aren’t canceled out are negative zero cubed when π‘Ÿ is equal to one and 𝑛 cubed when π‘Ÿ is equal to 𝑛. But of course, zero cubed is just equal to zero. So, we’re left with the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ cubed minus π‘Ÿ minus one all cubed is equal to 𝑛 cubed.

Let’s now consider the sum of the expression on the left-hand side of our equation. The sum from π‘Ÿ equals one to 𝑛 of three π‘Ÿ squared minus three π‘Ÿ plus one. To help with simplifying this sum, we can recall the summation linearity property. This states that for constants πœ† one and πœ† two, the sum from π‘Ÿ equals one to 𝑛 of πœ† one π‘Ž π‘Ÿ plus πœ† two 𝑏 π‘Ÿ is equal to πœ† one multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ž π‘Ÿ plus πœ† two multiplied by the sum from π‘Ÿ equals one to 𝑛 of 𝑏 π‘Ÿ. Applying this property, we have three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared minus three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ plus the sum from π‘Ÿ equals one to 𝑛 of one.

At this point, we can recall two other standard results for the sum of series which can be assumed. Firstly, the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ is equal to 𝑛 multiplied by 𝑛 plus one over two. And secondly, the sum from π‘Ÿ equals one to 𝑛 of a constant 𝛼 is equal to 𝛼 multiplied by 𝑛. Applying both these results, we have three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared minus three 𝑛 multiplied by 𝑛 plus one over two plus 𝑛. And that final 𝑛 comes from multiplying the constant one by 𝑛.

So, we found an expression for the sum of the series on the right-hand side and an expression for the sum of the series on the left-hand side, which are equal to one another. Everything in these two expressions is now in terms of 𝑛, apart from three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared. But the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is what we want to find an expression for. So, by equating these two expressions, we’ll then be able to rearrange to find what we’re looking for.

Equating the two expressions then, we have three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared minus three 𝑛 multiplied by 𝑛 plus one over two plus 𝑛 is equal to 𝑛 cubed. We’ll now clear some room to work through the algebraic simplification. To isolate the term we want, we add three 𝑛 multiplied by 𝑛 plus one over two to each side of the equation, and we also subtract 𝑛, giving three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared equals 𝑛 cubed plus three 𝑛 multiplied by 𝑛 plus one over two minus 𝑛.

To combine all of the terms on the left-hand side, we write them all with a common denominator of two, giving two 𝑛 cubed plus three 𝑛 multiplied by 𝑛 plus one minus two 𝑛 all over two. Distributing the parentheses in the center and then combining like terms, the right-hand side becomes two 𝑛 cubed plus three 𝑛 squared plus 𝑛 over two. Next, we factor the expression in the numerator by 𝑛 to give three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is equal to 𝑛 multiplied by two 𝑛 squared plus three 𝑛 plus one all over two.

In the next step, we’ll factor the quadratic two 𝑛 squared plus three 𝑛 plus one. It’s equal to 𝑛 plus one multiplied by two 𝑛 plus one. So, we have that three multiplied by the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one all over two. The final step is to divide both sides of the equation by three, giving the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one over six.

So, we found an algebraic expression for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared; it’s equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one over six. Remember, we should commit this result to memory. But if we needed to share its derivation, we recall that it comes from considering the binomial expansion of π‘Ÿ minus one cubed, which can then be rearranged to π‘Ÿ cubed minus π‘Ÿ minus one cubed is equal to three π‘Ÿ squared minus three π‘Ÿ plus one.

We then consider the sum from π‘Ÿ equals one to 𝑛 of the expressions on each side of this equation, recalling the standard results for the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ and the sum from π‘Ÿ equals one to 𝑛 of a constant. We found that the sum from π‘Ÿ equals one to 𝑛 of π‘Ÿ squared is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one over six.

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