Equal Coefficients and Unknown Constants

Transcript

0:00Introduction

Hello everyone! Okay, a small surprise to start with. Take three plus two x, and raise it to the power nine. Now find two coefficients: the coefficient of x cubed, and the coefficient of x to the power four. The first one is four lakh, eighty nine thousand, eight hundred and eighty eight. And the second one? Exactly the same number! Is that just luck? Or could we have predicted it? Hold that thought. Today, the question tells us something about the coefficients, and we work backwards to find the unknown constants inside. Questions like this come up often in JEE, so let's begin.

0:41The method: coefficients become equations

So, first recall the general term. T r plus one, capital T with a small r plus one written below, equals n C r, with the small n at the top left of a capital C, times the first term to the power n minus r, times the second term to the power r. And r runs zero, one, up to n; the three dots mean and so on. Okay? Here n is a positive integer, and we always count from r equals zero: the first term has r equals zero. Right? Now the method. Step one: the question gives you some coefficients. Write each one in terms of the unknowns, such as k or n. Keep every part: the n C r, the powers, and the number that comes with x. With me? Now tell me. If the question has two unknowns, how many equations do we need? Two, yes, exactly! Each condition in the question gives one equation. One unknown needs one equation; two unknowns need two. Clear? And step three, a small trick. Divide one equation by the other, or sometimes subtract them. The big factors, like large powers of three, cancel, and a simple equation is left. Got it?

1:58Back to the puzzle: equal coefficients in (3+2x)^9

Now, back to our puzzle: three plus two x, in round brackets, to the power nine. Using the general term, what is the coefficient of x to the power r? Yes, exactly! Nine C r, times three to the power nine minus r, times two to the power r. Let me call it c r: a small c, with r written below. Oh no, careful here! The second term is two x, so the two to the power r belongs to the coefficient. Don't drop it! Okay? So for x cubed, r is three. Nine C three is eighty four, three to the power six is seven hundred and twenty nine, and two cubed is eight. The small dot means times. Multiply them: four lakh, eighty nine thousand, eight hundred and eighty eight. Right? For x to the power four, r is four. One hundred and twenty six, times two hundred and forty three, times sixteen. Again, four lakh, eighty nine thousand, eight hundred and eighty eight. Confirmed: exactly equal. With me? But why? Let's divide neighbours, the trick from the last lesson. For three plus two x, to the power n, c r plus one, over c r, equals n minus r over r plus one, times two over three. The two over three comes from two x over three: the second number over the first. Clear? Now think. When are two neighbouring coefficients exactly equal? When the ratio is one, correct! Cross multiply: three times, r plus one, equals two times, n minus r. Notice, this works for any n. Got it? Now put n equals nine. Three r plus three equals eighteen minus two r. The double arrow you see on screen means, which gives. So five r equals fifteen, and r equals three. So the coefficients of x cubed and x to the power four had to be equal. No luck at all! Right?

4:03Example 1: equal x^2 and x^3 coefficients in (3+kx)^9

Alright, example one. The coefficients of x squared and x cubed in three plus k x, to the power nine, are equal. Find k, where k is not equal to zero; that is the equals sign with a slash through it. Pause the video and try it. Done? Okay, let's check. Same pattern as our puzzle. For x squared, r is two: nine C two, times three to the power seven, times k squared. For x cubed, r is three: nine C three, times three to the power six, times k cubed. Okay? The question says these are equal, so we set them equal. One condition, one equation, one unknown. Right? Now, the important question. Can we cancel k squared from both sides? Yes, but only because k is not zero! Look, k equals zero makes both coefficients zero, so equal too. That is why the question says k is not zero. So divide both sides by three to the power six, times k squared. On the left, nine C two is thirty six, and three to the power seven over three to the power six leaves three. On the right, nine C three is eighty four, and one k remains. So thirty six times three equals eighty four k. Clear? Thirty six times three is one hundred and eight. So k equals one hundred and eight over eighty four, which is nine over seven. See how the big powers of three just cancelled? Got it?

5:39Example 2: (1+ax)^n starts 1 + 8x + 24x^2, find a and n

Now a harder one, with two unknowns. The first three terms of one plus a x, to the power n, are one, eight x, and twenty four x squared. Find a and n. Pause the video and try it. Hint: two unknowns, so make two equations. Done? Okay, let's check. The first term is one to the power n, just one, so nothing new there. Second term: T two equals n C one, times a x. Now n C one is n, so n times a equals eight. That is equation one. Okay? Now the third term, whose coefficient is twenty four. So n C two times a equals twenty four, right? No, no, careful! That is the trap. In T three, the second term, a x, is squared, so we get a squared times x squared. So the coefficient is n C two times a squared: n times n minus one, over two, times a squared, equals twenty four. That is equation two. With me? Now, a neat move. Split the left side as n minus one, times a, over two, times n times a. And n times a is eight, from equation one! So n minus one, times a, over two, times eight, equals twenty four. Right? Eight over two is four, and twenty four over four is six. So n minus one, times a, equals six. Clear? Now look at the two equations. Here n times a is eight, and n minus one, times a, is six. Can you see a quick way to get a? Yes, subtract them! Here n times a, minus n minus one times a, leaves just one a. And eight minus six is two. So a equals two. Then n times two is eight, so n equals four. Two equations fixed two unknowns, just as we said. Got it? And always check. One plus two x, to the power four, is one, plus eight x, plus twenty four x squared, and so on; that is what the three dots mean. It matches perfectly. Okay?

7:58Quick revision: one equation per condition, then divide

So, quick revision. One: write each coefficient fully: the n C r, the powers, and the number with x. Two: one condition gives one equation; two unknowns need two. Three: divide or subtract the equations, but cancel only what is not zero. Four: in three plus two x, to the power n, c r equals c r plus one exactly when three times, r plus one, equals two times, n minus r. Now go and try the practice problems on this page. All the best!