Introduction to Binomial Expansion

BINOMIAL EXPRESSION

An algebraic expression containing two terms is called a binomial expression.

For example, (a + b), (2x – 3y), $\left(x+\frac{1}{y}\right),\left(x+\frac{3}{x}\right),\left(\frac{2}{x}-\frac{1}{x^2}\right)$ etc. are binomial expressions.

BINOMIAL THEOREM FOR POSITIVE INDEX

Such formula by which any power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. For a positive integer n , the expansion is given by

(a+x)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n–1 x + ⁿC₂ a^n-2 x² + . . . + ⁿC_r a^n–r x^r + . . . + ⁿC_nxⁿ = $\sum _{r=0}^n{}^n{C}_r{a}^{n-r}{x}^r$.

where ⁿC₀ , ⁿC₁ , ⁿC₂ , . . . , ⁿC_n are called Binomial co-efficients. Similarly

(a – x)ⁿ = ⁿC₀aⁿ – ⁿC₁a^n–1 x + ⁿC₂ a^n-2 x² – . . . + (–1)^{r n}C_r a^n–r x^r + . . . +(–1)^{n n}C_nxⁿ

i.e. (a – x)ⁿ = $\sum _{r=0}^n{\left(-1\right)}^r{}^n{C}_r{a}^{n-r}{x}^r$

Replacing a = 1, we get

(1 + x)ⁿ = ⁿC₀ +ⁿC₁x+ⁿC₂x² + . . . + ⁿC_r x^r + . . . + ⁿC_nxⁿ

and (1 – x)ⁿ = ⁿC₀ –ⁿC₁x+ⁿC₂x² – . . . + (–1)^{r n}C_r x^r + . . . +(–1)^{n n}C_nxⁿ

Observations:

$\bullet$There are (n+1) terms in the expansion of (a +x)ⁿ.

$\bullet$Sum of powers of x and a in each term in the expansion of (a +x)ⁿ is constant and equal to n.

$\bullet$The general term in the expansion of ( a+x)ⁿ is (r+1)^th term given as T_r+1 = ⁿC_r a^n-r x^r

$\bullet$The p^th term from the end = ( n –p + 2)^th term from the beginning .

$\bullet$Coefficient of x^r in expansion of (a + x)ⁿ is ⁿC_r a^{n - r} x^r.

^$\bullet$nC_x = ⁿC_y $\Rightarrow$ x = y or x + y = n.

$\bullet$In the expansion of (a + x)ⁿ and (a –x)ⁿ, x^r occurs in (r + 1)^th term.

Illustration 1: If the coefficients of the second, third and fourth terms in the expansion of (1 + x)ⁿ are in A.P., show that n = 7.

Solution: According to the question ⁿC₁ × ⁿC₂ × ⁿC₃ are in A.P.

$\frac{2n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{6}$

n² – 9n + 14 = 0 $\Rightarrow$ (n – 2)(n – 7) = 0 Þ n = 2 or 7

Since the symbol ⁿC₃ demands that n should be $⩾$ 3

n cannot be 2, $\therefore$ n = 7 only.

Illustration 2: Find the

(i) last digit (ii) last two digit (iii) last three digit of 17²⁵⁶.

Solution: 17²⁵⁶ = 289¹²⁸ = (290 –1)¹²⁸

= ¹²⁸C₀(290)¹²⁸ –¹²⁸C₁(290)¹²⁷ + ………..+ ¹²⁸C₁₂₆(290)² –¹²⁸C₁₂₇(290)+1

= 1000m + ¹²⁸C₂(290)² –¹²⁸C₁(290) + 1

= 1000m + $\frac{128\times 127}{2}$x(290)² –$\frac{128\times 290}{1}$ + 1 = 1000m + 683527680 + 1

Hence the last digit is 1. Last two digits is 81. Last three digit is 681.

Illustration 3: Find the coefficient of (i) x⁷ in ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$, (ii) and x^–7 in ${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$. Find the relation between a and b if these coefficients are equal.

Solution : The general term in ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}={}^{11}{C}_r{(a{x}^2)}^{11-r}{\left(\frac{1}{bx}\right)}^r$

= ${}^{11}{C}_r\frac{a^{11-r}}{b^r}{x}^{22-3r}$

If in this term power of x is 7, then 22 – 3r = 7 $\Rightarrow$ r = 5

$\therefore$ coefficient of x7 = ${}^{11}{C}_5\frac{a^6}{b^5}$ …(1)

The general term in ${\left(ax-\frac{1}{b{x}^2}\right)}^{11}={(-1)}^r{}^{11}{C}_r{(ax)}^{11-r}{\left(\frac{1}{b{x}^2}\right)}^r$

= ${(-1)}^r{}^{11}{C}_r\frac{a^{11-r}}{b^r}{x}^{11-3r}$

If in this term power of x is –7, then 11 – 3r = –7 $\therefore$ r = 6

$\therefore$ coefficient of x^–7 = (–1)⁶ ${}^{11}{C}_6\frac{a^{11-6}}{b^6}={}^{11}{C}_5\frac{a^5}{b^6}$

If these two coefficient are equal, then ${}^{11}{C}_5\frac{a^6}{b^5}={}^{11}{C}_5\frac{a^5}{b^6}$

$\Rightarrow$ $a^6{b}^6=a^5{b}^5\Rightarrow a^5{b}^5(ab-1)=0\Rightarrow ab=1(a\ne 0,b\ne 0)$

MIDDLE TERM

There are two cases

(a) When n is even

Clearly in this case we have only one middle term namely T_{n/2 + 1}. Thus middle term in the expansion of (a + x)ⁿ will be ⁿC_n/2 a^n/2x^n/2 term.

(b) When n is odd

Clearly in this case we have two middle terms namely $T_{\frac{n+1}{2}}and{T}_{\frac{n+3}{2}}$. That means the middle terms in the expansion of (a +x)ⁿ are ${}^n{C}_{\frac{n-1}{2}}.a^{\frac{n+1}{2}}.x^{\frac{n-1}{2}}$ and ${}^n{C}_{\frac{n+1}{2}}.a^{\frac{n-1}{2}}.x^{\frac{n+1}{2}}$.

Illustration 4: Find the middle term in the expansion of ${\left(3x-\frac{x^3}{6}\right)}^9$.

Solution: There will be two middle terms as n = 9 is an odd number. The middle terms will be ${\left(\frac{9+1}{2}\right)}^{th}$ and ${\left(\frac{9+3}{2}\right)}^{th}$ terms.

t₅ = ⁹C₄(3x)^{5${\left(-\frac{x^3}{6}\right)}^4=\frac{189}{8}{x}^{17}$}

t₆ = ⁹C₅(3x)^{4${\left(-\frac{x^3}{6}\right)}^5=-\frac{21}{16}{x}^{19}$}.

GREATEST BINOMIAL COEFFICIENT

In the binomial expansion of (1 + x)ⁿ , when n is even, the greatest binomial coefficient is given by ⁿC_n/2.

Similarly if n be odd, the greatest binomial coefficient will be ${}^n{C}_{\frac{{}_{n+1}}{2}}and{}^n{C}_{\frac{{}_{n-1}}{2}},bothbeingequal.$

NUMERICALY GREATEST TERM

If t_r and t_{r + 1} be the r^th and (r + 1)^th term in the expansion of (1 + x)ⁿ, then

$\frac{t_{r+1}}{t_r}=\frac{{}^n{C}_r{x}^r}{{}^n{C}_{r-1}{x}^{n-r}}=\frac{n-r+1}{r}$x.

Let numerically, t_{r + 1} be the greatest term in the above expansion. Then t_{r + 1} $⩾$ t_r

or $\frac{t_{r+1}}{t_r}$ $⩾$ 1 $\Rightarrow$ $\frac{n-r+1}{r}$|x| $⩾$ 1

$\Rightarrow$ r $⩽$ $\frac{\left(n+1\right)|x|}{\left(1+|x|\right)}$ ……(2)

Now shifting values of n and x in (2), we get r $⩽$ m + f or r $⩽$ m

Where m is a positive integer, f is a fraction such that 0 $⩽$ f < 1.

Now if f = 0 then t_{m + 1} and t_m both the terms will be numerically equal and greatest while if f $\ne$ 0, then t_{m +1} is the greatest term of the binomial expansion.

i.e. to find the greatest term (numerically) in the expansion of (1 + x)ⁿ.

(i) Calculate m = $\frac{(n+1)|x|}{(1+|x|)}$.

(ii) If m is integer, then t_m and t_{m + 1} are equal and are greatest term.

(iii) If m is not integer, then t_{[m] + 1} is the greatest term (where [.] denotes the greatest integer function).

Illustration 5: Find the value of the greatest term in the expansion of $\sqrt{3}{\left(1+\frac{1}{\sqrt{3}}\right)}^{20}$.

Solution: Since $\frac{t_{r+1}}{t_r}=\frac{\sqrt{3}\cdot {}^{20}{C}_r{\left(\frac{1}{\sqrt{3}}\right)}^r}{\sqrt{3}\cdot {}^{20}{C}_{r-1}{\left(\frac{1}{\sqrt{3}}\right)}^{r-1}}=\frac{{}^{20}{C}_r}{{}^{20}{C}_{r-1}}\cdot \frac{1}{\sqrt{3}}=\frac{21-r}{r}\cdot \frac{1}{\sqrt{3}}$

$t_{r+1}\ge t_r$ if only 21 – r$⩾$$r\sqrt{3}$

if only r $⩽$ $\frac{21}{\sqrt{3}+1}=\frac{21(\sqrt{3}-1)}{2}$ = 7.686

Hence t₁ < t₂ < t₃ < t₄ < t₅ < t₆ < t₇ < t₈ > t₉ > t₁₀

Hence t₈ is the greatest term and its value is $\sqrt{3}\cdot {}^{20}{C}_7{\left(\frac{1}{\sqrt{3}}\right)}^7$.

= $\sqrt{3}\cdot {}^{20}{C}_7\frac{1}{{\left(\sqrt{3}\right)}^6}={}^{20}{C}_7\frac{1}{3^3}=\frac{25840}{9}=2871.11$