APPENDIX : BINOMIAL THEOREM

Binomial theorem is something that has been known to mathematicians since many centuries ago. In this introduction, we’ll trace the origins of this theorem to the coefficients we obtain when we expand any binomial term raised to an integral power.

Consider the following expansions, which can be verified by direct multiplication:

(

^x+y

)

^{0 = 1}

Do you notice anything special about these expansions, in particular, any general rule or trend these expansions follow that might enable us to expand

(

^x+y

)

ⁿ directly for a general n ? First of all, notice that the number of terms in each expansion is one more than the power of the binomial term. For example,

(

^x+y

)

⁵ has 6 terms.

However, mathematicians long back also realized another important fact, namely, the relation between the coefficients obtained upon expansion. To see what this relation is, let us write the coefficients in the following

‘triangular’ pattern:

Do you observe any relation between the various coefficients. If not, consider this same arrangement in a

The ‘rule’ for constructing this triangular pattern should be pretty obvious now. All edge-numbers are 1. Any other is obtained by adding the number directly above and to the left with the number directly above and to the right, as in Fig-34 Extending this process gives us all the ‘binomial coefficients.’ This geometrical arrangement of the binomial coefficients in a triangle is called Pascal’s triangle. The figure below shows a Pascal triangle containing the coefficients upto n = 15.

1 13 78 286 715 1287 1716 1716 1287 715 286 78 13

1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 corresponds toⁿC_i), we will immediately understand that (1) is equivalent to

1 1

1

n n n

i i i

C = ⁻C₋ + ⁻C

which, as we already know from the last chapter on P & C, is true.

Section - 1 BINOMIAL THEOREM, POSITIVE INTEGRAL INDEXSection - 1

Let us now consider more formally the binomial theorem. We need to expand

(

^x+y

)

ⁿ, where x, y are two arbitrary quantities, but n is a positive integer.

In particular, if we write

(

^x+y

) (

^{n = x+y}

)(

^x+y

)(

^x+y

) (

^{... x+y}

) (

^{n times}

)

^{... (1)}

we need to find out the coefficient of x y^{i j}. Note that i+ j must always equal n, so that we can write a general term of the expansion (without the coefficient) as x y^{r n r−} so that 0≤ ≤r n.

Now, to find the coefficient of x y^{r n r−} , note that we need the quantity x, r times, while y is needed n – r times.

Thus, in (1), x y^{r n r−} will be formed whenever x is ‘contributed’ by r of the binomial terms, while y is ‘contributed’

by the remaining n - r of the binomial terms. For example, in the expansion of

(

^x+y

)

^{5, to form x y2 3, we}

need x from 2 terms and y from 3:

(

^x+y

)

^{5 =}

(

^x+y

)(

^x+y

)(

^x+y

)(

^x+y

)(

^x+y

)

take x

take y One of the ways to form x y^{2 3}

Fig - 23

How many ways are there to form x y^{2 3}? In other words, how many times will x y^{2 3} be formed? The number of times x y^{2 3} is formed is what is the coefficient of x y^{2 3}. That number, which would be immediately obvious to the alert reader, is simply⁵C₂. Why? Because this is the number of ways in which we can select any 2 binomial terms from 5. These 2 terms will contribute x. The remaining will automatically contribute y.

In the general case of

(

^x+y

)

ⁿ, we see that the coefficient of x y^{r n r−} would be ⁿC_r. (which is infact the same as ⁿC_{n r−} ). Thus, the general binomial expansion is

(

x+y

)

ⁿ= ⁿC x0 ⁿ+ ⁿC1 x^n-1y + ⁿC x2 ^n-2 y²+...+ⁿC y_n ⁿ

( )

0

n n n n r r

r r

x y C x ⁻ y

=

⇒ + =

∑

The coefficients ⁿC_i are called the binomial coefficients, for a reason that should now be obvious.

Note that the

( )

^{i+1 th} coefficient in this expansion is ⁿC_i, which now explains the relation

, 1 1, 1, 1

n i n i n i

T ₊ =T₋ +T_{− +}

we observed in the Pascal triangle; this relation simply corresponds to

1 1

1

n n n

i i i

C = ⁻C₋ + ⁻C

Also, the binomial coefficients of terms equidistant from the beginning and the end are equal, because we have

n n

Since we have (n + 1) terms in the general expansion, we see that if n is even, there will be an odd number of terms, and thus there will be only one middle term, which would be ⁿC_{n/ 2}x^{n/ 2}y^{n/ 2}. For example,

On the other hand, if n is odd , then there will be an even number of terms in the expansion, and thus there will be two middle terms, namely

1 1 1 1

Find the middle term(s) in the expansion of 1 9

x x

 − 

 

 

Solution: Since n = 9, there will be 10 terms in the expansion, which means that there will be 2 middle terms in the expansion, the 5^th and the 6^th:

Example – 2

Is there any term in the expansion of

2 10

Solution: The general term in the expansion is

Thus, for the term that is independent of x, we have

20 5 0

Solution: We have already evaluated this sum in the chapter on P & C. That approach was as follows: this sum basically counts the number of all sub-groups of a set of size n; this can also be counted by focusing on each element of the set, which has two corresponding choices - you either include it into your sub-group or you don’t, which means that the total number of ways to form sub-groups is 2 × 2 × 2 .... n times = 2ⁿ. The sum of the binomial coefficients therefore equals 2ⁿ.

Here, we evaluate the same sum using a binomial approach. Consider the following expansion:

(

1+x

)

ⁿ = ⁿC0+ⁿC x1 +ⁿC x2 ² +ⁿC x3 ³+ +... ⁿC x_n ⁿ If we put x = 1, we simply obtain

0 1 2

2ⁿ =ⁿC +ⁿC +ⁿC +...ⁿC_n

Thus, the same result is obtainable from both a combinatorial and a binomial approach.

We can also derive another useful result by putting x = –1 in the above relation, so that we obtain

This states the sum of the even-numbered coefficients is equal to the sum of the odd-numbered coefficients. Can you prove this using a combinatorial approach?

As an exercise, prove the following relations:

ⁿC₀2ⁿ+ⁿC₁2ⁿ⁻¹+ⁿC₂2ⁿ⁻²+...ⁿC_n =3ⁿ no role to play in this part.

To find the greatest coefficient, consider the following ratio:

( ) ( )

Similarly,

we see that for odd n, the two middle coefficients are the greatest. This can be verified by considering the following expansion:

( )

^{5 5 4 3 2 2 3 4 5}

The two middle coefficients are the greatest for odd

In this case therefore, the greatest coefficient is the single middle coefficient

2 n

Cn. Lets verify this again:

( )

^{6 6 5 4 2 3 3 2 4 5 6}

The single middle coefficient is the greatest for even

6 15 20 15 6

n

x+y = +x x y+ x y + x y + x y + xy +y

!

(b) To find the greatest term, we must also consider x and y. We again follow the approach of

⇒ are the two greatest terms

Now, if

(

^{n 1}

)

^y

Example – 5

How will you expand the multinomial expression

(

x¹+ + +x^{2 ...} xm

)

ⁿ?

Solution: We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient)

x x x₁ⁿ¹ ₂ⁿ² ₃ⁿ³....x_m^nm ... (1) where the various powers must always sum to n (why?).

i.e.,

1 2 3 ... _m

n + + + +n n n =n

Now, to evaluate the coefficient of the term in (1), we consider the multinomial expression in expanded form: the number of ways in which this can be done.

First select those n₁ multinomials that will contribute x₁ : this can be done in

1

n

Cn ways. Now, from the remaining

(

n n− 1

)

multinomials, select those n₂ multinomials that will contribute x₂ : this can be done in^{( 1)}

2

n n

Cn

− ways. Continuing this process, we see that the number of ways to get

1from 1, 2

This is what is known as the general multinomial coefficient. The multinomial expansion can now be written compactly as

where the summation is carried out over all possible combinations of the n_i's such that

∑

n_i =n^.

Find the coefficient of x⁴ in the expansion of

10

Solution: From the previous example, the general term in the expansion will be

3 Thus, the (total) coefficient of x⁴ is

0 1 2

10! 10! 10!

( 2) ( 2) ( 2)

6! 4! 0! − +3! 6!1! − +0!8! 2! − = – 1290 (verify)

Section - 2 DIFFERENTIATION & INTEGRATION TECHNIQUESSection - 1

The techniques of calculus enable us to sum a lot of series involving binomial coefficients. This is the subject of this section.

Suppose that we have to evaluate the sum S given by

1 2 2 3 3 ...

n n n n

S = C + C + C + +n Cn

From now on, to avoid clutter, we’ll write ⁿC_r as simply C_r, where the upper index n should be understood to be present. Thus,

1 2 2 ... ⁿ _n S =C + C + + C =∑rC_r

This series can be generated using a manipulation involving differentiation, as follows:

Consider the binomial expansion

2

0 1 2

(1+x)ⁿ =C +C x+C x +...+C x_n ⁿ If we differentiate both sides with respect to x, look at what we’ll obtain:

1 2 1

1 2 3

(1 )ⁿ 2 3 ... _n ⁿ

n +x ⁻ =C + C x+ C x + +nC x ⁻ Now, all that remains is to substitute x = 1, upon which we obtain:

1 Thus, we have evaluated another interesting sum.

Suppose that we now wish to evaluate S₁ given by

1 2

The alert reader would immediately realize that integration needs to be applied here. How exactly to do so is now described. Consider again the expansion.

2

0 1 2

(1+x)ⁿ =C +C x+C x +...+C x_n ⁿ

If we integrate this with respect to x, between some limits say a to b, we obtain

1 2 3 1

To generate the sum S₁, a little thought will show that we need to use a = 0, b = 1, so that we obtain

Try some other values for a and b and hence generate other series on your own. Be as varied as you can in choosing these limits.

Example – 7 Find the sum S given by

2 2 2 2

1 2 3

1 2 3 .... _n

S = ⋅ + ⋅C C + ⋅C + + ⋅n C

Solution: We have to plan an approach wherein we are able to generate r² with C_r. We can generate one r with every C_r, as we did earlier, and which is now repeated here:

2

0 1 2

(1+x)ⁿ =C +C x C x+ +...+C x_n ⁿ Differentiating both sides with respect to x, we have

1 2 1

1 2 3

(1 )ⁿ 2 3 ... _n ⁿ

n +x ⁻ = +C C x+ C x + +nC x ⁻

Now we have reached the stage where we have an r with every C_r. We need to think how to get the other r. If we differentiate once again, we’ll have r(r – 1) with every C_r instead of r²(understand this point carefully). To ‘make-up’ for the power that falls one short of the required value, we simply multiply by x on both sides of the relation above to obtain:

1 2 3

1 2 3

(1 )ⁿ 2 3 .... _n ⁿ

nx +x ⁻ =C x+ C x + C x + +nC x It should be evident now that the next step is differentiation:

n n( −1) (1x +x)ⁿ⁻²+n(1+x)ⁿ⁻¹ =C₁+ ⋅2² C x₂ + ⋅3² C x₃ ² +....+n²⋅C x_n ⁿ⁻¹ Now we simply substitute x = 1 to obtain

2 1 2 2 2

1 2 3

( 1) 2ⁿ 2ⁿ 2 3 ... _n

n n − ⋅ ⁻ + ⋅n ⁻ =C + ⋅C + ⋅C + + n ⋅C The required sum S is thus

2 1

( 1) 2ⁿ 2ⁿ S =n n− ⋅ ⁻ + ⋅n ^{− = ⋅n 2n−2}

{

^{(n− +1) 2}

}

=n n( + ⋅1) 2ⁿ⁻²

Example – 8

Solution: The first sum contains only the even-numbered binomial coefficients, while the second contains only odd-numbered ones. Recall that we have already evaluated the sum S given by

1

Thus, if we determine S₁, S₂ is automatically determined, and vice-versa. Let us try to determine S₁ first.

(a) Consider again the general expansion

2

0 1 2

(1+x)ⁿ =C +C x+C x + +.... C x_n ⁿ

Integrating with respect to x, we have (we have not yet decided the limits)

1 2 3 1

Since we are trying to determine S₁ which contains only the even-numbered terms, we have to choose the limits of integration such that the odd-numbered terms vanish. This is easily achievable by setting a = – 1 and b = 1 (understand this carefully). Thus, we have

1

Section - 3 MISCELLANEOUS TECHNIQUESSection - 1

Not all questions can be subjected to the method(s) described earlier. For example, consider the sum S given by

0 1 1 2 2 3 ... _n 1 _n

S =C C +C C +C C + +C C₋

Let us first go through a combinatorial approach, using the observation that C₀ =C C_n, ₁=C_n−1 and so on, so that S can be rewritten as

1 1 2 2 3 ... 1

n n n n

S =C C +C C₋ +C ₋C + +C C

Consider a general term of this sum, which is of the form C_{n r−} C_r+1. We can think of this as the number of ways of selecting (n – r) boys from a group of n boys and (r + 1) girls from a group of n girls. The total number of people we are thus selecting is (n− + + = +r) (r 1) (n 1). Therefore, S represents the total number of ways of selecting (n + 1) people out of a group of 2n, so that S is simply ²ⁿC_n+1.

Now to a binomial approach. This will involve generating the general term C C_{r r+1} somehow, which is the same as C_{n r−} C_r+1. Consider the general expansion of (1+x)ⁿ.

2

0 1 2

(1+x)ⁿ =C +C x+C x +...+C x_n ⁿ ...(1) We have to have the terms C C C C_{n 1}, _{n−1 2} and so on, which suggests that we write (1) twice, but in the second expansion we reverse the terms, multiply, and see what terms contain the (combinations of) coefficients we require.

(1 + x = C + C x + C x + ...+ C x)ⁿ 0 1 2 ⁿ 2

n

(1 + x = C x + C)ⁿ n ⁿ n – 1x + Cⁿ n – 2xⁿ + ...+ C x + C1 0

–1 – 2

Multiplying, we find on the left hand side we have (1+x)²ⁿ, while on the right hand side, the terms containing the (combinations of) coefficients we want will always be of the form ( )xⁿ⁺¹, that is, the power of x will be (n + 1). No other terms will contain xⁿ⁺¹, verify this for yourself. Thus, the sum C C_{n 1}+C C_{n−1 2}+...+C C_{1 n} is actually the total coefficient of xⁿ⁺¹ on the right hand side, and from the left hand side we know that the coefficient of xⁿ⁺¹ would be simply ²ⁿC_n+1. Thus, S= ²ⁿC_n+1

A very similar approach could have been:

(1 + ) = x ⁿ C0 + C + 1x C x2 + ... + C xⁿ

2

n

1 + 1 x

n

= C₀ + C₁1

x + C₂ 1

x² + ... + C_n1 xⁿ

Thus, S = Coefficient of x in (1 ) 1 ¹ Find the sum S given by

2 2 2 2

0 1 2 ... _n S =C +C +C + +C

Solution: Note that S can be rewritten as

0 _n 1 _n 1 2 _n 2 ... _n 0

S =C C +C C₋ +C C ₋ + +C C

Using a combinatorial approach, the sum should be immediately obvious to the alert reader as

2n

Cn. In brief, this is because the right hand side represents, as an example, the total number of ways of selecting n people from a group of n boys and n girls, etc.

Now, we discuss the binomial expansion approach:

2 3 Find the sum S given by

1 2

0 1 2 ...

n n n n r

S = C + ⁺C + ⁺ C + + ⁺ Cr

Solution: We have already evaluated this sum in P & C; here we’ll use a binomial approach. Note that ^{n r+} C_r =Coeff. of xⁿ in (1+x)^{n r+}

⇒ ^{∑n r+Cr = ∑}

(

^{Coeff. of xn in (1+x)n r+}

)

= Coeff. of xⁿ in∑ +(1 x)^{n r+}

Thus,

Coeff. of ⁿ in (1 )ⁿ (1 )ⁿ 1 ... (1 )^{n r} S = x  +x + +x ⁺ + + +x ⁺ 

^{=Coeff. of xn in (1+x)n}

{

^{1 (1+ + + +x) (1 x)2+... (1+ +x)r}

^{(1 )}

{

^{(1 ) 1 1}

}

Coeff. of in

n r

n x x

x x

+ + + −

=

^{=Coeff. of xn+1in (1}

{

^{+x)n r+ +1− +(1 x)n}

}

=^{n r+ +1}C_r

⇒ S =^{n r+ +1}Cr, which is the same result we obtained in P & C.

TRY YOURSELF - I

Q 1. Find the sum of all the rational terms in the expansion of

(

^{31/ 4+41/ 3}

)

^12.

Q. 2 Prove that three consecutive terms in a binomial expansion can never be in G.P.

Q. 3* (a) Show that the integral part of

(

^{5 5 11+}

)

^{67 is even}

(b) Show that the integral part of

(

^{8 3 7+}

)

^{n is odd}

Q. 4 Use the binomial theorem to show that 7¹⁰³ when divided by 5 leaves a remainder 3.

Q. 5 Find the coefficient of x³⁰¹ in the expansion of

500 499 2 498 500

Q. 7* Use the binomial theorem to show that 32³²³² when divided by 7 leaves the remainder 4.

Q. 8 Prove the following relations:

(a) ⁿC₀⋅²ⁿC_n−ⁿC₁⋅²ⁿ⁻¹C_n+ⁿC₂⋅ ²ⁿ⁻²C_n−...+ⁿC_n( 1)− ⋅^{n n}C_n =1

Q. 9 Find the sums of the following series.

(a) C₀−2C₁+3C₂−4C₃+... (b)* ^{0 1 2 3 ...}

3 4 5 6

C −C +C −C +

Q. 10 Using the binomial theorem, show that

0 Q. 11 Find the sum of the series

0

Section - 4 BINOMIAL THEOREM, RATIONAL INDEXSection - 1

In the previous section, we discussed the expansion of (x+y)ⁿ, where n is a natural number. We’ll extend that discussion to a more general scenario now. In particular, we’ll consider the expansion of (1+x)ⁿ, where n is a rational number and | x | < 1. Note that any binomial of the form (a b+ )ⁿ can be reduced to this form.: The general binomial theorem states that

2 3

That is, there are an infinite number of terms in the expansion with the general term given by

1

For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains an infinite number of terms, we have:

(

^1+x

)

^{n =a0+a x1 +a x2 2+a x3 3+ +... a xn n+ ∞...}

Putting x = 0 gives a₀ = 1. Now differentiating once gives

(

¹

)

^{n 1 1 2 2 3 3 2 ...}

n +x ⁻ = +a a x+ a x + ∞ Putting x = 0 gives a₁ = n.

Proceeding in this way, we find that the r^th coefficient is given by

(

¹

)(

^{2 ...}

) (

¹

)

Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because T_r+1 becomes ⁿC x_r ^r, and the expansion terminates for r>n. For the general T_r+1, we obviously cannot use ⁿC_r since that is defined only for natural n.

One very important point that we are emphasizing again is that the general expansion holds only for | | 1x < .

Let us denote the genral binomial coefficient by V_r. Thus, we have

Let us discuss some particularly interesting expansions. In all cases, | | 1:x <

(1) (1+x)⁻¹: Since n= −1, we see that

Solution: (a) We have

The coefficient of xⁿ in this binomial expansion (note: the power is now a positive integer) would be ( 1)− ⁿ⋅³ⁿC_n.

Example – 12

Find the magnitude of the greatest term in the expansion of

(

^{1 5 y−}

)

^{−2 / 7 for 1}

y=8.

Solution: Let us first do the general case: what is the greatest term in the expansion of (1+x)ⁿ, where n is an arbitrary rational number. We have,

1

Now, let us find the conditions for which this ratio exceeds 1. We have

1

r r

T₊ ≥ T

⇒ ⁿ_r^{+ − ≥1 1} _{| |}¹_x ...(1)

For this particular problem, (1) becomes Find the sum of the series

2 2.5 2 5 8

Solution: We need to determine n and x. For that, we can compare the terms of this series with the corresponding terms in the following general expansion.

^{(1 ) 1 ( 1) 2 ( 1) ( 2) 3 ...}

Solving for n and x from these two equations, we get ²

n= −3 and ¹

Example – 14

Find the sum of the series

1 1 1

Solution: Before solving this problem, ponder a moment over the following fact:

In the expansion of (1+x)ⁿ, if x << 1, that is, if x is much smaller than 1, then the expansion can be approximated as

(1+x)ⁿ ≈ +1 nx

since all higher order terms can be neglected due to the small magnitude of x.

Coming to the problem, note that if b << a, i.e, if ^b ₁

Thus, the sum S of the series is (to a good approximation)

1

Evaluate 99^3/2 correct to four decimal places.

Solution: We have

^{993 / 2 =}

(

^{100 1−}

)

^{3 / 2}

( )

^{3 / 2}

3 / 2

100 1 0.01

= ⋅ −

( )

^{3 / 2}

1000 1 0.01

= ⋅ −

( ) ( )

²

3 1

3 2 2

1000 1 0.01 0.01 ...

2 2!

 ⋅ 

 

= ⋅ − ⋅ + ⋅ − 

 

 

3 3

1000 1 ...

200 80000

 

=  − + −  1000 15 0.0375 ...

= − + −

= 985.0375

Note that we only considered the first three terms of the expansion because the higher order terms would not have had any effect on the answer upto the fourth decimal place.

In document Correspondence Sample Study Material on Binomial Theorem (Page 62-85)