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IIT-JEE 2009

Binomial Theorem (Part-I)

Author: anuraagchandra

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If n is a positive integer:
- $(x+y)^n=^nC_0x^{n-0}y^0+^nC_1x^{n-1}y^1+^nC_2x^{n-2}y^2+\cdots +^nC_nx^{n-n}y^n$
  $=\overset {n}{\underset {r=0}{\sum}}^nC_rx^{n-r}y^r$
- $(x+y)^n=x^n+^nC_1x^{n-1}y^1+^nC_2x^{n-2}y^2+\cdots +y^n \quad [\therefore ^nC_0=1, ^nC_n=n]$
- $(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+\cdots +^nC_nx^n=\overset {n}{\underset {r=0}{\sum}}^nC_rx^r$
- $(1-x)^n=^nC_0-^nC_1x+^nC_2x^2-\cdots +(-1)n.^nC_nx^n=\overset {n}{\underset {r=0}{\sum}}(-1)^r.^nC_rx^r$
- - In the expansion of , coefficient of $x^r=^nC_{r-1}$
  - In the expansion of , co-efficient of
  - Number of terms in the expansion of
How to directly write any positive integral power of a binomial expression:
- $(\alpha+\beta)^5=\alpha^5+5\alpha^4\beta+10\alpha^3\beta^2+10\alpha^2\beta^3+5\alpha\beta^4+\beta^5(\alpha+\beta)^5$
  $= \alpha^5+5\alpha^4\beta^1+\frac {5*4}{2}\alpha^3\beta^2=10$
  $+\frac {10*3}{3}\alpha^2\beta^3+\frac {10*2}{4}\alpha\beta^4+\frac {5*1}{5}\beta^5$
  $=10 \qquad \qquad =5 \qquad =1$
  $=\alpha^5+5\alpha^4\beta+10\alpha^3\beta^2+10\alpha^2\beta^3+5\alpha\beta^4+\beta^5$
- $(\alpha-\beta)^5= \alpha^5-5\alpha^4\beta+10\alpha^2\beta^2-10\alpha\beta^3+5\alpha\beta^4-\beta^5$
- $(\alpha+\beta)^6+\alpha^6+6\alpha^5\beta+\frac {6*5}{2}\alpha^4\beta^2+\frac {15*4}{3}\alpha^3\beta^3+\frac {20*3}{4}\alpha^2\beta^4+\frac {15*2}{5}\alpha\beta^5+\frac {6*1}{6}\beta^6$
  $= \alpha^6+6\alpha^5\beta+15\alpha^4\beta^2+20\alpha^3\beta^3+15\alpha^2\beta^4+6\alpha\beta^5+\beta^6$
- $(x+y)^n=x^n+\frac {n}{1}x^{n-1}y+\frac {n(n-1)}{2}x^{n-2}y^2+\cdots +y^n$
- $(1+x)^n=1+\frac {n}{1}x+\frac {n(n-1)}{2}x^2+\cdots +x^n$
- If n is a negative integer or a fraction
  $(1+x)^n=1+\frac {n}{1}x+\frac {n(n-1)}{2}x^2+\frac {n(n-1)(n-2)}{3}x^3+\cdots to \quad \infty \quad where -1 < x < 1$
rth Term:
- In the expansion of , rth term is given by $t_r=^nC_{r-1}X^{n-r+1}y^{r-1}$
  Example: In the expansion of $\Big(x-\frac {3}{x^2} \Big)^{20},t_7(7^{th}term)$
  $=^{20}C_6x^{20-6}\Big(-\frac {3}{x^2}\Big)^6=^{20}C_6x^{14}\Big(-\frac {3}{x^2}\Big)^6$
- $r^{th}$ term from end in the expansion of $(x+y)^n=r^{th}$ term from beginning in the expansion of .
- $r^{th}$ term from end in the expansion of $(x+y)^n=(n-r+2)^{th}$ term from beginning in
Co-efficient of rth term:
In the expansion of , coefficient of $r^{th}$ term $=^nC_{r-1}$
Example: In the expansion of $\Big(x-\frac {3}{x^2}\Big)^{20},$ coefficient of $5^{th}term =^{20}C_4$
Middle term:
Number of terms in the expansion of
- If power n is even, then middle term is $\Big(\frac {n}{2}+1\Big)^{th}$ term.
  Number of middle term = 1
- If power n is odd, then there will be two middle terms.
  They are $\frac {n+1}{2}$ th term and $\frac {n+3}{2}$ th term.
- Middle term has greatest coefficient. If there are two middle terms, then their coefficients are equal.
  Example:
  - In the expansion of , co-efficient of terms are
    
    i.e. 1, 4, 6, 4, 1
    Greatest coefficient = 6 = coeff. of 3rd term = coeff. of middle term
  - In the expansion of , coefficients of terms are
    i.e. 1, 5, 10, 10, 5, 1
    greatest coefficients = 10, 10
    Here middle terms are 3rd and 4th terms. Their coefficients are greatest and equal.