The General Term and Terms from the End

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Find the fifth term of two x squared minus one over x, the whole thing to the power eight. Most students start at once: fifth term, so eight C five, right? Hmm, is it really? By the end of this video you will know, and you will never mix up a term number again. Questions like this come up often in JEE, so let's begin.

0:29The general term: counting starts at r=0

So, first, recall the expansion. For a positive integer n, x plus a, to the power n, is the sum of n C r, x to the power n minus r, a to the power r. The big Greek letter sigma means add up, for r from zero to n. And n C r is the C with a small n at the top left and r at the bottom right. Okay? Now tell me. The very first term: which value of r gives it? Yes, exactly, r equals zero! The first term, n C zero, x to the power n, has r equal to zero. The second term, n C one, x to the power n minus one, times a, has r equal to one. So the term number is always one more than r. Right? So, the term with a to the power r is term number r plus one, because counting starts at zero. On screen it is written as r plus one in brackets, followed by t h. Clear? Now the key line. We call this term T r plus one: capital T, with r plus one written small at the bottom right. T r plus one 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. This is the general term. Write it down. Got it? Now, with a minus sign inside, x minus a, to the power n, just replace a with minus a. So T r plus one gets an extra minus one to the power r in front. Even r gives a plus sign, odd r gives a minus sign. With me?

2:00The rth term: the biggest trap

Now, this is where most marks are lost. Suppose a question asks for the r-th term. You just write n C r, x to the power n minus r, a to the power r, right? No, no, careful! For the r-th term, the counting index is one less: r minus one. So T r equals n C r minus one, x to the power n minus r plus one, a to the power r minus one. Okay? And n C r, x to the power n minus r, a to the power r is T r plus one, one term later. Most errors in this chapter come from mixing up r and r plus one. Right? So, here is the habit toppers use: always write the general term as T r plus one first, then solve for r. Keep this in mind for our puzzle. Clear?

2:45Terms from the end

Alright, now terms from the end. Since r runs from zero to n, x plus a, to the power n, has n plus one terms, and the last one is T n plus one. Okay? So the last term is the first from the end. Then the second from the end: which term is it from the beginning? T n, yes, exactly! Just one step back. With me? So for the k-th term from the end, start at the last term and step back k minus one times. n plus one, minus k, plus one: that is n minus k plus two. So the k-th term from the end is T n minus k plus two. Clear? And here is a shortcut: flip the binomial. The k-th term from the end of x plus a, to the power n, is also the k-th term of a plus x, to the power n. Got it?

3:37Ratio of neighbouring terms

One last result. Put the two neighbours side by side. T r plus one is n C r, x to the power n minus r, a to the power r. And T r is n C r minus one, x to the power n minus r plus one, a to the power r minus one. Now divide the first by the second. What is left of the powers of x and a? Just a over x. Correct! And n C r over n C r minus one simplifies to n minus r plus one, over r. The dot means times. So T r plus one over T r equals n minus r plus one over r, times a over x. Okay?

4:15Example 1: 5th term of (2x^2 - 1/x)^8

Alright, now our puzzle from the start. Find the fifth term of two x squared minus one over x, the whole thing to the power eight. Pause the video and try it. Done? Okay, let's check. Fifth term means r plus one equals five, so r is four. Eight C four, not eight C five! Puzzle solved. Right? So T five equals eight C four, times two x squared to the power four, times minus one over x, in brackets, to the power four. With me? Now quick. Minus one over x, to the power four: is this term positive or negative? Positive, because the power is even. Correct! Eight C four is seventy, two x squared to the power four is sixteen x to the power eight, and minus one over x to the power four is x to the power minus four. The dots mean times. Okay? Seventy times sixteen is one thousand one hundred and twenty, and x to the power eight times x to the power minus four, gives x to the power four. So the fifth term is one thousand one hundred and twenty x to the power four. Perfect!

5:26Example 2: 4th term from the end of (x - 2/x)^9

Now a harder one. Find the fourth term from the end of x minus two over x, the whole thing to the power nine. Pause the video and try it. Done? Okay, let's check. Power nine means ten terms. Fourth from the end: ten minus four is six, so the sixth term, right? No, no, careful! Count backwards. T ten is the first from the end, then T nine, T eight, and T seven is the fourth. Okay? And our formula agrees: n minus k plus two is nine minus four plus two, that is seven. So r plus one is seven, and r is six. Clear? So T seven equals nine C six, times x to the power nine minus six, that is x cubed, times minus two over x, in brackets, to the power six. With me? Now you work out minus two over x, to the power six. Sixty four over x to the power six, positive, since the power is even. Correct! Nine C six is eighty four, and the power of x is three minus six, that is minus three. Eighty four times sixty four: so the answer is five thousand three hundred and seventy six over x cubed. Very good! And a quick check with the shortcut: flip the binomial. The fourth term of minus two over x plus x, to the power nine, is nine C three, times minus two over x to the power six, times x cubed. Since nine C three equals nine C six, it is the same term. That's it!

6:56Quick revision: T(r+1), the rth term, terms from the end

So, quick revision. The general term: T r plus one equals n C r, x to the power n minus r, a to the power r, counting from r equals zero. For the r-th term, use r minus one, and for x minus a, multiply by minus one to the power r. And the k-th term from the end is T n minus k plus two, or the k-th term of a plus x, to the power n. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!