Daniel has been exploring the relationship between Pascal’s triangle and the binomial expansion. He has noticed that each row of Pascal’s triangle can be used to determine the coefficients of the binomial expansion of 𝑥 plus 𝑦 to the power of 𝑛, as shown in the figure. For example, the fifth row of Pascal’s triangle can be used to determine the coefficient of the expansion of 𝑥 plus 𝑦 to the power of four.
This is a two-part question. The first part says by calculating the next row of Pascal’s triangle, find the coefficients of the expansion of 𝑥 plus 𝑦 to the power of six. So this is Pascal’s triangle named after the French mathematician Blaise Pascal. As stated in the question, the numbers in each row of the triangle give the coefficients in the binomial expansion of 𝑥 plus 𝑦 to the power of 𝑛.
For example, the numbers one, three, three, one in the fourth row of Pascal’s triangle give the coefficients in the expansion of 𝑥 plus 𝑦 cubed. You can of course confirm this yourself by expanding 𝑥 plus 𝑦 cubed fully and then grouping like terms. But Pascal’s triangle gives us a shortcut for doing each of these expansions.
We’re asked first of all to calculate the next row of pascal’s triangle in order to find the coefficients of the expansion of 𝑥 plus 𝑦 to the power of six. So let’s look at this triangle and see what patterns we can identify. First, we can see that the number of values in each row increases by one each time. So there’ll be seven values in the seventh row of the triangle. Next, we see that each row of the triangle begins and ends with the number one. So we can fill in the two numbers on the end of our row.
Next, if we look diagonally at the triangle, we can see an increasing pattern — one, two, three, four, five — going in each direction. To continue this pattern, the second and penultimate numbers in our row of the triangle must both be six.
Now, let’s look at how to fill in the remaining numbers in our row. And to do this, we need to look at the relationship between each row and the row above it. If we look at the number six in the fifth row of the triangle, we can see that it’s the sum of the two numbers above it — three and three. Similarly, if we look at the number 10 in the sixth row of the triangle, we can see that it is also the sum of the two numbers above it — six and four. In fact, this pattern repeats throughout the triangle and this is how we calculate numbers in the next row.
The third number in our row is found by adding five and 10 together to give 15. The next number is found by adding 10 and 10, giving 20. The final number in our row is found by adding 10 and five together, giving 15. You could also fill in this number by spotting that each row of Pascal’s triangle is symmetrical. So we found the next row of Pascal’s triangle. And it tells us that the coefficients in the expansion of 𝑥 plus 𝑦 to the power of six will be one six, 15, 20, 15, six, one.
If we want to look what the full expansion would look like, we know that each term has decreasing powers of 𝑥 from six to zero and increasing powers of 𝑦 from zero to six. So the expansion would be 𝑥 to the power of six plus six 𝑥 to the five 𝑦 plus 15𝑥 to the four 𝑦 squared plus 20𝑥 cubed 𝑦 cubed plus 15𝑥 squared 𝑦 to the four plus six 𝑥𝑦 to the five plus 𝑦 to the power of six. So that is our answer to the first part of the question. And I’m now going to delete some of this working to make way for the second part.
Daniel now wants to calculate the coefficients for each of the terms of the expansion two 𝑥 plus 𝑦 to the power of four. By substituting two 𝑥 into the expression above or otherwise, calculate all of the coefficients of the expansion.
So what we’ve been given is an extract from Pascal’s triangle. It’s the fifth row which tells us the coefficients in the expansion of 𝑥 plus 𝑦 to the power of four. We however want to know about the expansion of two 𝑥 plus 𝑦 to the power of four. The question suggests that one way to do this is to substitute two 𝑥 into the expression above. So that’s how we will go about it. So I’ve used the expansion of 𝑥 plus 𝑦 to the power of four. But everywhere that I had 𝑥 previously, I’ve now replaced it with two 𝑥.
It’s really important that you notice that the two and the 𝑥 are treated in exactly the same way. For example, they are both raised to the power of four in the first term. The first term will be two to the power of four multiplied by 𝑥 to the power of four, which is 16𝑥 to the power of four. A really common mistake is to forget that that two is also raised to the power of four and therefore think that the first term should just be two 𝑥 to the four. It isn’t. It’s 16𝑥 to the four.
In exactly the same way, the second term should be four multiplied by two cubed multiplied by 𝑥 cubed multiplied by 𝑦. Two cubed is equal to eight and four multiplied by eight is 32. So the second term is 32𝑥 cubed 𝑦. The common mistake would again be to forget that the two is also raised to the power of three and think that the second term is just four multiplied by two multiplied by 𝑥 cubed multiplied by 𝑦, giving eight 𝑥 cubed 𝑦.
Continuing the expansion, we must again remember that in the third term the two is squared this time. So we have six multiplied by two squared multiplied by 𝑥 squared multiplied by 𝑦 squared. You’re less likely to make mistakes with the final two terms as you don’t have to raise two to a power other than one in either of them.
Simplifying the coefficients for the final three terms, we have six multiplied by four which is 24𝑥 squared 𝑦 squared plus four multiplied by two which is eight 𝑥𝑦 cubed and then plus 𝑦 to the power of four.
The coefficients then in the expansion of two 𝑥 plus 𝑦 to the power of four are 16, 32, 24, eight, and one. Remember you must treat that two in exactly the same way as 𝑥 and raise it to the same power as 𝑥 every time.