Find the middle term in the expansion of (3/x + x/3)^10

Question

Accepted Answer

Hey everyone here we are asked to determine the seventh term of the expansion of the given expression. Where we have the quantity of X divided by five plus five divided by x raised to the 12th power. Now, before we begin solving this problem, we first need to recall the binomial theorem which tells us that for any positive integer N we have the quantity of A plus B to the end. Power is equal to the sum from R is equal to zero to N. Of N R times A. To the power of n minus R times B to the power of R. So now we just need to start identifying the values for each of these variables in the formula in relation to our given expression. So we can start with our value for our while looking at our formula. We can see that our First term in the expansion begins when R is equal to zero. So this tells us that our value for our is always going to be one less than the term we're looking for. So in this case we are looking for the seventh term. So we just take 7 -1, telling us our value for our is six. Now we can start finding the rest of the variables. We can begin with the value for a. Well A is just the left hand side term from our given expression. So A will be X divided by five, which leaves us with be being five divided by X. And lastly we need to find the value for N. Well N is just the power to which the binomial is raised to. So our end in this case is going to be 12. And with this information we can go ahead and start finding our seventh term. So beginning with the coefficient here, Our top most term is N. So we will have 12. And then the lower term is our so we will have six times a which is again X divided by five. And we raise this term to the power of N minus R which is just 12 minus six. Then we multiply this by B which is again five divided by X. And raise it to the power of our which is six. Now we can just start simplifying. So first just keeping the coefficient the same again we have 12 and six. And then for our next term our X divided by five is raised to the power of 12 minus six which simplifies to the quantity of X divided by five raised to the six power. And then our last term we can just rewrite, we have five divided by X raised to the power of six. And with this we can again start simplifying. So we have the coefficient as 12 and six. And then with our next term distributing the sixth power we have X to the sixth power divided by five to the sixth power times the next quantity is five to the six power divided by X to the sixth power. So now we can make some clear cancelation here we see that these X. X terms cancel as well as the five to the power of six. Both cancel since they are in the numerator and the denominator. So this leaves us with just our coefficient. Now if we recall for this coefficient in the formatting N. R. This is equal to n factorial divided by R. Factorial times the quantity of n minus R factorial. So with this formula in mind we can go ahead and write out our coefficient. So in the numerator we have 12 factorial which is divided by the R factorial which is six factorial times the quantity Of N - R. Which again is 12 -6 factorial. And now we can just start simplifying. So keeping the numerator the same. For now we have 12 factorial divided by six factorial times 12 -6 gives us six factorial. So having a six factorial in the denominator here we can expand our numerator which is 12 factorial until we also have a six factorial in the numerator. So doing this we have 12 times 11 times times nine times eight times seven times six factorial. All divided by six factorial times six factorial. And so from this we can clearly see that one of these six factorial in the denominator cancels out with the six factorial in the numerator. So this leaves us with in the numerator again we have 12 times 11 times 10 times nine times eight times seven. All divided by six factorial. Which expanding we have in the denominator six times five times four times three times two times one. And now we can make some more easy simplifications. So we see we have a four times three in the denominator while we have a 12 in the numerator. So these terms cancel. We also have a five times two in the denominator. Well we have a 10 in the numerator. So these terms also cancel and that looks like that's all these simplifications we can make. So multiplying out the numerator we have 5544. And in the denominator we just have six times one giving us six. And now dividing these terms we get the seventh term which is 924. This is our answer here for choice D Thanks for watching. I hope you found this video helpful and I'll see you again next time

Find the middle term in the expansion of (3/x + x/3)^10

Watch next