Numerically Greatest Term

10. Numerically Greatest term in the binomial expansion: (x + y )n

When the terms in a binomial expansion are written, each of them contains the binomial coefficients and coefficients of variables x and y, if any.

Therefore, there is some term in which the number resulting from the product of binomial coefficients and coefficients of variables in the binomial terms is greatest in value.

In this lesson, our task is to find which term contains such greatest number or numerically the greatest term.

Recall that to find a term is to find r. So, let’s get on with the task of finding r.

Eg. Find the numerically greatest term in the expansion of (2 – 3x)7, when x = 1

Solution: In order to understand the concept the numerically greatest term clearly, let us write all the terms in the given binomial expansion (2 – 3x)7 , as it contains not too many terms and also since the coefficient of the variables x and y namely 2 and 3 are small in value.
(2 – 3x)7 = 7c0 (2)7 + 7c1 (2)6(-3x)1 + 7c2 (2)5 (-3x)2 + 7c3 (2)4 (-3x)3 + 7c4 (2)3 (-3x)4 + 7c5 (2)2 (-3x)5 + 7c6 (2)1 (-3x)6 + 7c7 (-3x)7 ……………..(1)

Use ncr to write the values of the binomial coefficients.
7c0 = 1, 7c1 = 7, 7c2 = 21, 7c3 = 35, 7c4 = 35, 7c5 = 21, 7c6 = 7, 7c7 = 1

And the values of powers of 2 are
27 = 128, 26 = 64, 25 = 32, 24 = 16, 23 = 8, 22 = 4, 21 = 2

And the values of powers of 3 are

31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243, 36 = 729, 37 = 2187

Plug in the values of binomial coefficients, the values of powers of 2 and powers of 3 in the respective places of the terms in (1) above, we see

(2 – 3x)7 = 1(128) + 7 (64)(-3) + 21(32)(9) + 35(16)(-27) + 35(8)(81) + 21(4)(-243) + 7(2)(729) + 1(-2187)………………….(2)

(we have chosen here to not write x, since the question is about finding the term in which the number is the greatest in value. So we do not need x)

In (2) above, we see that the term having the greatest numerical value is 35(8)(81)

{ here we must understand what numerical value means. Consider 2 and -2. The two numbers are different and therefore not equal. But their numerical values is 2 each and therefore same. Numerical value of a number is not concerned with its negative or positive sign. It is just the number without the sign. In (2) above too, we must disregard in the terms the negative signs of the numbers resulting from the product of the binomial coefficients and the coefficients of the variables x and y. Only the numerical values, i.e., the numbers without signs must be compared}

Now, we clearly see and understand that there is some term in any binomial expansion in which the number resulting from the product of the binomial coefficients and the coefficient of the variables x and y is greatest in value.

The next question is how to find this term having the greatest numerical value in any binomial expansion without writing down all the terms in the tedious way discussed above. And of course, the above way of expanding the binomial expansion to locate the term with the greatest numerical value was just a way to illustrate the fact that some term will have the numerically greatest number. That’s not the way always.

Then what’s the right way?

You guessed it? No, Ok. Let me remind you once again.

To find a term having the greatest numerical value is same as finding what? Yes, finding r

In (2) above, we see that the fifth term contained the numerically greatest number. It was 35(8)(81).

It was the fifth term, T5. The numerical value in this fifth term was greater than the numerical values of every other term in the given binomial expansion above.

Now, we call this fifth term the general term in order to find r to arrive at it (without expanding the binomial expansion, of course)

Let’s denote this general term (containing the number greater in value than all the numbers in all of the other terms in the binomial expansion) as Tr + 1

Again, set Tr as any other term compared to which Tr + 1 is greater, i.e.,
Tr + 1 > Tr

See binomial expansion below:
(x + y)5 = x5 + 5x4y + 10x3 y2 +10x2y3 +5xy3 + y5 )

We see that in the third and the fourth terms, the number is 10 each, i.e. equal. So, sometimes two terms can have the same greatest numerical value. Therefore, let us reset it as

Tr + 1 Tr

The given binomial expansion is (2 – 3x)7, when x = 1
Now Tr + 1 = 7cr (2) n – r (3x)r

Tr = 7c r – 1 (2) n – ( r – 1 ) (3x)r – 1

Now, plug 1 in x and ignore the negative sign in -3x, since we want the greatest numerical value.

T r + 1 = 7cr (2) n – r (3)r
Tr = 7c r – 1 (2) n – ( r – 1 ) (3)r – 1
7cr (2) n – r (3)r        7c r – 1 (2) n – ( r – 1 ) (3)r – 1

Now
[ ( 7 – r + 1 )/ r ]. 3        2
24 – 3r       2r
24       5r
r       4.8
that means, the numerically greatest term is contained in the term in which r = 4 and this is the fifth term, i.e., T5