# Application of Binomial Theorem for Finding Integral Solutions

## Binomial Theorem

The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem.

(x + y)n = nC0 xn + nC1 xn-1 y + nC2 xn-2 y2……………..nCn yn

e.g:  (a+b)33C0a3 + 3C1a2b+3C2ab2 + 3C3b3

= a3 + 3a2b + 3ab2 +b3

Note

• The number of terms in the expansion of (x+ y)n is (n + 1) e. one or more than the index .
• The sum of the indices of x & y in each term is always n.

## General term of the expansion

The general term or the (r + 1)th the term in the expansion of (x + y)n is given by:

Tr+1 = nCr xn-r yr

Example 1: The co-efficient of x2 in the expansion of ${\left( {{x^2} + \frac{2}{x}} \right)^7}$ is

Solution:

General term of the expansion of (x+y)n= Tr+1 = nCr xn-r yr. comparing with the expression in the problem, here x = x, y = 2/x and n = 7. Therefore,

Tr+1 = 7Cr(x2)7-r(2/x)r

Or Tr+1 = 7Cr$\times$2r$\times$(x)14-3r.

Clearly, for r =4, the power of x becomes 2. Hence putting r =4 we get the required coefficient = 7C4$\times$24.

## Middle Terms

The  middle  term(s)  is  the  expansion  of  (x + y)n  is (are)  :

1. If n  is even, there  is  only  one  middle  term  which  is  given  by ;

T(n+2)/2 = nCn/2 . xn/2 . yn/2

1. If n  is  odd ,  there  are  two  middle  terms  which  are  :T(n+1)/2    &    T((n+1)/2)+1

Example 2: Find the middle term of the expansion (x + 1/x)8

Solution: Here n = 8 is even, hence the total number of terms in the given expansion is 9. Therefore, the middle term is the 5th term, i.e.

T5 = 8C4$\times$x8-4$\times$(1/X)4  = 8C4.

Example 3: The total number of terms in the expansion of (x+y)200 + (x-y)200 after simplification is

Solution:

(x + y)200 = 200C0 x200 + 200C1 x199 y + 200C2 x198 y2……………..200C200 y200 ………. Eq(1)

(x – y)200 = 200C0 x200 – 200C1 x199 y + 200C2 x198 y2……………..200C200 y200 ………… Eq(2)

The total number of terms in both the equations is 201 each.

Observe that in eq(2), the coefficients of terms having an odd power of x are negative I (what is ‘I’ here)sign but equal in magnitude to the corresponding coefficients of the terms in eq (1). And the number of such terms in both the equations is 100.

On adding both the equations, these terms get canceled. Hence the total number of terms in the expansion of (x+y)200 + (x-y)200 is 100.

Numerically greatest term

To  find  the  numerically  greatest  term  is  the  expansion  of  (1 + x)n, put the absolute value of x & find the largest value of r  consistent  with  the  inequality $\frac{{{T_{r + 1}}}}{{{T_r}}} \ge 1$

Example 4: Find the greatest term in the expansion of (3+2x)11 if x = 2/3.

Solution:

, putting n = 11 and x = 2/3 we get,

$\frac{{{T_{r + 1}}}}{{{T_r}}} = \left( {\frac{{n – r + 1}}{r}} \right)\left( {\frac{{2x}}{3}} \right)$ , putting n = 11 and x = 2/3 we get,

$\frac{{{T_{r + 1}}}}{{{T_r}}} = \left( {\frac{{12 – r}}{r}} \right){\left( {\frac{2}{3}} \right)^2}$

$= \left( {\frac{{12 – r}}{r}} \right)\left( {\frac{4}{9}} \right) = \frac{{48 – 4r}}{{9r}}$

Therefore

${T_{r + 1}} \ge {T_r}$, if, 48-4r $\ge$9r

Or, $r \le \frac{{48}}{{13}}$ as r has to be integer, the largest value of r = 3.

Hence the greatest term in the expansion for x = 2/3 is T4 = 11C3$\times$38$\times {\left( {\frac{4}{3}} \right)^3}$

## Important Points for expansion (x1 + x2 + x3 + ….. + xr)n

### Number of terms

The total number of terms in the above expansion is given by n+r-1Cr-1

Example 5: find the number of terms in the expansion (a + b + c + d)50

Solution:

This is like distributing 50 identical objects in 4 groups. So this can be done in (n+r-1)C (r-1) ways. This means it would be (50+4-1) C (4-1) = 53C3

Coefficient of $x_1^{{n_1}} \times x_2^{{n_2}} \times \ldots . \times x_r^{{n_r}}$ such that n1 +n2 + n3 + ….. + nr = n is given by: $\frac{{n!}}{{\left( {{n_1}!} \right)\left( {{n_2}!} \right)\left( {{n_3}!} \right) \ldots \ldots \left( {{n_r}!} \right)}}$

Example 6: Find the coefficient of a4b2c3din the expansion of (a + b + c + d)14.

Solution:

It is like distributing 14 people into 4 groups of sizes 2, 3, 4 & 5. Therefore, the required coefficient is $\frac{{14!}}{{2!3!4!5!}}$

Example 7: Find the coefficient of ${a^3}{b^4}{c^5}$ in the expansion of ${(bc + ca + ab)^6}$

Solution:

In this case, ${a^3}{b^4}{c^5} = {(ab)^x}{(bc)^y}{(ca)^z} = {a^{x + z}}.{b^{x + y}}.{c^{y + z}}$

z + x = 3, $x + y = 4,\,y + z = 5$;  $2(x + y + z) = 12$;  $x + y + z = 6$. Then $x = 1,y = 3,\,z = 2$

Therefore the coefficient of ${a^3}{b^4}{c^5}$ in the expansion of ${(bc + ca + ab)^6}$ = $\frac{{6!}}{{1!\,3!\,2!}} = 60$.

### Method for finding terms free from radical or rational terms in the expansion of ${({a^{1/p}} + {b^{1/q}})^N}\rlap{–} \vee a,\,b\, \in$prime numbers

Example 8: The number of integral terms in the expansion of ${{\left( \sqrt{3}+\sqrt[8]{5} \right)}^{256}}$ is

Solution:

${T_{r + 1}}{ = ^{256}}{C_r}\,.\,{3^{\frac{{256 – r}}{2}}}\,.\,{5^{\frac{r}{8}}}$

First term = $^{256}{C_0}\,\,{3^{128}}{5^0} = {\rm{integer}}$ and after eight terms, i.e., 9th term = $^{256}{C_8}{3^{124}}{.5^1} = {\rm{integer}}$.Continuing like this, we get an A.P., ${1^{{\rm{st}}}},\,{9^{th}}…….\,{257^{th}}$;  ${T_n} = a + (n – 1)\,d$

$\Rightarrow 257 = 1 + (n – 1)\,8$

$\Rightarrow$n=33

Example 9: The number of irrational terms in the expansion of ${{\left( \sqrt[8]{5}+\sqrt[6]{2} \right)}^{100}}$ is

Solution:

${T_{r + 1}}{ = ^{100}}{C_r}\,{5^{\frac{{100 – r}}{8}}}{.2^{\frac{r}{6}}}$

As 2 and 5 are co-prime. ${T_{r + 1}}$ will be rational if $100 – r$ is multiple of 8 and r is multiple of 6 also $0 \le r \le 100$

$\Rightarrow \,r = 0,\,6,\,12…….96$;

$\Rightarrow \,100 – r = 4,\,10,\,16…..100$………….. (i)

But  100-r is to be multiple of 8.

So, 100-r= 0, 8, 16, 24,……96 …..(ii)

Common terms in (i) and (ii) are 16, 40, 64, 88.

$\Rightarrow$ r = 84, 60, 36, 12 give rational terms

$\Rightarrow$ The number of irrational terms  = 101 – 4 = 97.

## Finding the integral or fractional part of the expansion

Example 10: The integral of ${(\sqrt 2 + 1)^6}$ will be

Solution:

Let ${(\sqrt 2 + 1)^6}$=I + f, where I is the integral part and f is the fractional part.

Also, let f’ be the complementary fraction of f, such that f + f’ = 1. {Both f and f’ are proper fraction i.e. fractions less than 1}

Note that, $\left( {\sqrt 2 – 1} \right) < 1$ or ${\left( {\sqrt 2 – 1} \right)^6} < 1$. Let f’=${\left( {\sqrt 2 – 1} \right)^6}$

Also, we know that, ${(x + y)^n} + {(x – y)^n} = 2({x^n}{ + ^n}{C_2}{x^{n – 2}}{y^2}{ + ^n}{C_4}{x^{n – 4}}{y^4} + …..)$

Or ${(\sqrt 2 + 1)^6} + {(\sqrt 2 – 1)^6}$ =I + f + f’

$= 2({(\sqrt 2 )^6}{ + ^6}{C_2}{(\sqrt 2 )^4}{(1)^2}{ + ^6}{C_4}{(\sqrt 2 )^2}{(1)^4}{ + ^6}{C_6}{(\sqrt 2 )^0}{(1)^6})$

= $2(8 + 15 \times 4 + 30 + 1) = 198$

As f +f’ =1, the required integral part is 198-1 = 197.