Mixed Problems: Set 1 (Terms and Coefficients)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I give you three numbers: seven hundred and twenty nine, seven thousand two hundred and ninety, and thirty thousand three hundred and seventy five. And I tell you, these are the first three terms of x plus a, the whole thing to the power n. Can you find x, a and n? Three unknowns, hiding inside powers. Looks scary, right? By the end of this class, you will crack it in under two minutes. This is Set one of our mixed problems, on terms and coefficients, and questions like these come up often in JEE Main. So, let's begin.

0:42Set 1: terms and coefficients

So, here is what Set one holds. Fifteen solved problems, all in the JEE Main pattern. Short, sharp questions, okay? Some ask for a given power of x, or the constant term. Some ask for a middle term, or a term counted from the end. Right? Others ask for the number of rational terms, the greatest coefficients, or a condition on neighbouring terms. All mixed, just like the real exam. Clear? Each one should take you one to two minutes. So attempt the whole set first, and only then read the solutions. Today: two tools, then two problems from the set, one easy, one harder. With me?

1:24Tool 1: the general term and a power of x

Tool number one, from Chapter two: the general term. For n a positive integer, in x plus a, to the power n, the general term is T r plus one: capital T, with r plus one written small at the bottom right. It equals n C r, x to the power n minus r, a to the power r, for r equals zero, one, and so on up to n; the three dots mean and so on. And n C r is the C with a small n at the top left and r at the bottom right. Remember, it is term number r plus one, not term number r, because counting starts at r equals zero. Okay? Now, for any binomial, write it like this: n C r, times the first term to the power n minus r, times the second term to the power r. Each term goes inside its own bracket. Right? Then collect the powers of x into one power, and set it equal to k, the power the question wants. Solve for r. That's it! Clear? Now tell me. The term independent of x, the constant term: what power of x does it have? Zero, yes, exactly! No x left at all. So for the term independent of x, the constant term, just set the power equal to zero. Got it? But careful. The r you get must be a whole number from zero to n; the square brackets mean both ends are included. If r comes out as a fraction, or outside that range, there is no such term, and the coefficient is zero. With me? And one more habit: keep the signs and the numbers inside the brackets. Drop a minus sign, and the answer goes wrong even with the right r.

3:11Tool 2: the ratio of neighbouring terms

Okay? Tool number two, also from Chapter two: the ratio of neighbouring terms. In x plus a, to the power n, T r plus one over T r equals n minus r plus one, over r, times a over x. The dot on screen means times. When a question gives actual terms, this ratio is your best friend. Clear? So, put r equals one. What is T two over T one? Yes! n minus one plus one is just n, and r is one. So T two over T one is n times a, over x. Okay? Now put r equals two. T three over T two is n minus one, over two, times a over x. Look at that two at the bottom: it is the r. Don't lose it! Got it? Now the clever part. Divide one ratio by the other. What cancels? The a over x cancels, exactly! And only n is left: one equation, one unknown. That's the trick. With me?

4:19Example 1: term independent of x in (x^3 - 2/x^2)^10

Now, the first problem from the set, an easy one. Find the term independent of x in x cubed minus two over x squared, the whole bracket to the power ten. Pause the video, and try it first. Done? Okay, let's check. Tool one. T r plus one equals ten C r, times x cubed to the power ten minus r, times minus two over x squared, to the power r. See, the minus sign stays inside the bracket. Okay? Now collect the power of x. From x cubed: three times ten minus r. From one over x squared: minus two r. Total: thirty minus five r. Right? So, which value of r makes this power zero? Six, yes! Thirty minus five r equals zero, and the double arrow on screen means, so, r equals six. And r equals six means the seventh term, T seven, not the sixth. Clear? Now, there is a minus sign inside the bracket. So the answer must be negative, right? No, no, careful! Minus two to the power six is plus sixty four, because six is even. And ten C six is two hundred and ten. So, two hundred and ten times sixty four: thirteen thousand four hundred and forty. Positive! Perfect!

5:48Example 2: find x, a, n from terms 729, 7290, 30375

Now, remember our puzzle from the start? Let's finish it. The first three terms of x plus a, to the power n, are seven hundred and twenty nine, seven thousand two hundred and ninety, and thirty thousand three hundred and seventy five. Find x, a and n. Pause the video, and try it. Hint: use tool two. Done? Okay, let's check. T two over T one is n times a, over x. And seven thousand two hundred and ninety over seven hundred and twenty nine is exactly ten. Okay? Now T three over T two. Put r equals two: n minus one, times a, over x, and that's all, right? No, no, wait! The r at the bottom! It is n minus one, over two, times a over x. And thirty thousand three hundred and seventy five over seven thousand two hundred and ninety is twenty five over six. Clear? Now divide. The a over x cancels. On the left, n times two over n minus one: two n over n minus one. On the right, ten times six over twenty five is twelve over five. Halve both sides: n over n minus one is six over five. So n is six. Got it? Now back to the first ratio. With n equal to six, six a over x is ten, so a over x is ten over six, that is five over three. With me? And T one is x to the power six, which is seven hundred and twenty nine. So x is three, taking x greater than zero, as the note in brackets says, because minus three to the power six is also seven hundred and twenty nine. Then a is five over three times three: five. Okay? Always check. T two is six, times three to the power five, times five: seven thousand two hundred and ninety. It matches! So x is three, a is five, and n is six. Puzzle solved, in under two minutes. Perfect!

8:07Quick revision: power of x, signs, ratio of neighbour terms

So, quick revision. One: for a power of x, write T r plus one, collect the power, and solve for r. Two: watch the signs; minus two to the power r is positive when r is even. Three: when terms are given, divide the neighbouring ratios, a over x cancels, and you get n. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!