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Binomial Theorem (Part-II)

by Anurag Chandra

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  1. Greatest Term:
    • In a binomial expansion greatest term means numerically greatest term.
      \therefore greatest term in (x-y)^n(x,y >0)= greatest term in (x+y)^n
    • If rth term is the greatest term, then \frac {t_r}{t_{r+1}}\underline >1 and \frac {t_r}{t_{r-1}}\underline >1
      But if there is only one greatest term, then for rth term to be greatest term \frac {t_r}{t_{r+1}}\underline >1 and \frac {t_r}{t_{r-1}}\underline >1

    • In order to find the greatest term in the expansion of (1+x)^n, calculate r=\frac {(n+1)|x|}{|x|+1}
      • If r is not an integer, and k be the integer just less than r, then (r+1) th term is the greatest term.
      • If r is an integer, then rth and (r+1) th terms are the greatest terms and they are equal.



    • (1+x)^n=^nC_0+^nC_1x+^nC_2x^2+\cdots +^nC_nx^n \qquad \qquad \cdots (1)
      Putting x = 1, we get
      2^n=^nC_0+^nC_1+^nC_2+^nC_3+\cdots +^nC_n \qquad \qquad \cdots(2)
      Putting x = â€"1 in (1), we get
      0=^nC_0-^nC_1+^nC_2-^nC_3+\cdots +(-1)^n.^nC_n \qquad \qquad \cdots (3)
      (2)+(3)\Rightarrow 2n=2[^nC_0+^nC_2+^nC_4+\cdots ]
      \Rightarrow 2^{n-1}=^nC_0+^nC_2+^nC_4+ \qquad \qquad \cdots (4)
      Here we cannot write the last term on R.H.S. unless it is known whether n is odd or even.
      (2)-(3)\Rightarrow 2n=2[^nC_1+^nC_3+^nC_5+\cdots]
      \Rightarrow 2^{n-1}=^nC_1+^nC_3+^nC_5+ \qquad \qquad \cdots (5)
      Thus,
      (i) ^nC_0+^nC_1+^nC_2+\cdots +^nC_n=2n
      (ii) ^nC_0- ^nC_1+^nC_2-\cdots +(-1)^{n-1}^nC_n=0
      (iii) ^nC_0+^nC_2+^nC_4+\cdots = 2^{n-1}
      (iv) ^nC_1-^nC_3+^nC_5\cdots =2^{n-1}

    • Note:
      (i) ^{10}C_0+^{10}C_2+^{10}C_4+\cdots +^{10}C_{10}=2^{10-1}=2^9
      (ii) ^{10}C_1+^{10}C_3+^{10}C_5+\cdots +^{10}C_9=2^{10-1}=2^9
      (iii) ^{11}C_0+^{11}C_2+^{11}C_3+\cdots +^{11}C_{10}=2^{11-1}=2^{10}
      (iv) ^{11}C_1+^{11}C_3+^{11}C_5+\cdots +^{11}C_{11}=2^{11-1}= 2^{10}
      Here last term is nCn if in each term upper and lower suffices are both even or both odd.
      Last term is nCnâ€"1 if in each term one of lower and upper suffices is even and other is odd.

    • If Cr stands for ^nC_r , then
      • C_0C_r+C_1C_{r+1}+C_2C_{r+2}+\cdots +C_{n-r}C_n=^{2n}C_{n-r}

      • C_0^2+C_1^2+C_2^2+\cdots C_n^2=^{2n}C_n



  2. If \sqrt {x} is an irrational number whose square is an integer and y is an integer, such that 0<y-\sqrt {x}<1 , then
    • For even positive integer n,
      If (y+\sqrt {x})^n = p+f, where p\in I and 0<f<1 and (y-\sqrt {x})^n=f_1 , where 0<f1<1
      Then f+f_1=1 and p is an even integer
      Also (p+f)(1-f)=(y^2-x)^n

    • For odd positive integer n and 0<\sqrt {x}-y<1
      if (\sqrt {x}+y)^n=\rho +f , where \rho \in I,0<f<1 and (\sqrt {x}-y)^n=f_1 , where 0<f_1<1
      Then f=f_1 and p is an odd integer (\rho +f)f= (x-y^2)^n


  3. Properties of ^nC_r : ^nC_r is also denoted by \Big(\overset {n}{\underset {r}{}}\Big)

    (a) ^nC_{r-1}+^nC_r=^{n+1}C_r
    (b) ^nC_x=^nC_y\Rightarrow x=y \quad or \quad x+y=n
    © ^nC_r=^nC_{n-r}
    (d) r.^nC_r=n.^{n-1}C_{r-1}
    (e) ^nC_r=\frac {n}{r}.^{n-1}C_{r-1}
    (f) \frac {^nC_r}{r+1}= \frac {^{n+1}C_{r+1}}{n+1}
    (g) \frac {^nC_r}{^nC_{r-1}}= \frac {n-r+1}{r}
    (h) \frac {^nC_r}{^nC_{r+1}}= \frac {r+1}{n-r}
    (i) If n is even, ^nC_r is greatest when r= \frac {n}{2} .
    If n is odd, ^nC_r is greatest when r=\frac {n+1}{2}or \frac {n-1}{2}

    • In any expansion containing x, term independent of x means coefficient of x^0

      .

    • Coeff. of x^r in x^k\Big(x-\frac {3}{x^2}\Big)^n= coeff.of \quad x^{r-k}in \Big(x- \frac {3}{x^2}\Big)^n

    • (1+3x-5x^4)^n may be taken as a_0+a_1x+a_2x^2+\cdots +a_{4n}x^{4n}, where some of a_0,a_1,\cdots ,a_{4n} , may be zero.



  4. Multinomial theorem:
    The expansion of (x_1+x_2+x_3+x_4+\cdots +x_m)^n
    when n \in N can be obtained by means of the binomial theorem.
    The general term in the expansion of (x_1+x_2+x_3+x_4+\cdots +x_m)^n = \frac {n}{\alpha_1\alpha_2\cdots \alpha_m}x_1 \quad ^{\alpha 1}x_2 \quad ^{\alpha 2}\cdots x_m \quad ^{\alpha m}
    where \alpha_1,\alpha_2,\cdots ,\alpha_m are nonâ€"negative integers satisfying the condition \alpha_1+\alpha_2+\cdots +\alpha_m=n

    Proof: General term in the expansion [x_1+(x_2+\cdots +x_m)]^n
    =\frac {n}{\alpha_1|n - \alpha_1}x_1 \quad \alpha_1(x_2+x_3+\cdots +x_m)^{n-\alpha_1}\qquad \qquad \cdots (1)
    Again general term in the expansion of (x_2+x_3+\cdots +x_m)^{n-\alpha 1} =\frac {n-\alpha_1}{\alpha_2 |n-\alpha_1-\alpha_2}x_2\alpha_2.(x_3+x_4+\cdots +x_m)^{n-\alpha_1-\alpha_2} \qquad \cdots (2)
    Similarly general term in the expansion of (x_3+x_4+\cdots +x_m)^{n-\alpha_1-\alpha_2}= \frac {n-\alpha_1-\alpha_2}{\alpha_3|n-\alpha_1-\alpha_2-\alpha_3}x_3^{\alpha 3}
    (x_4+x_5+\cdots x_m)^{n-\alpha_1-\alpha_2-\alpha_3} \qquad \qquad \qquad \qquad \qquad \cdots (3)
    Proceeding in this way we can show that the general term in the expansion of (x_1+x_2+\cdots +x_m)^n

    =\frac {n}{\alpha_1|n-\alpha_1}.\frac {|n-\alpha_1}{|\alpha_2|n-\alpha_1-\alpha_2}.\frac {|n-\alpha_1-\alpha_2}{|\alpha_3|n-\alpha_1-\alpha_2-\alpha_3}.x_1^{\alpha_1}x_2^{\alpha_2}.x_3^{\alpha_3}\cdots x_m \quad ^{\alpha_m}

    where \alpha_1+\alpha_2+\cdots +\alpha_m=n and \alpha_1,\alpha_2,\cdots ,\alpha_m are nonâ€"negative integers.

    (a) General term in (a+b+c+d)^n=\frac {|n}{|p|q|r|s}a^p.b^q.c^r.d^s , where p+q+r+s=n;p,q,r,s,\in[\{0,1,2,\cdots ,n\}]
    (b) Number of terms in the expansion of (x_1+x_2+\cdots +x_m)^n = number of ways of distributing n identical things among m persons when each person can get zero or more things =m\quad + \quad n \quad - \quad ^1C_n \quad or \quad m \quad + \quad n \quad - \quad ^1C_m \quad - \quad 1
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    spiyr
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    suman sourabhSun, 04 Apr 2010 10:00:21 -0000

    what type of questions can be asked from the topic "Multinomial theorem"

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anuraagchandra
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About: I am an engineer by profession, designing software applications and leading project teams in software development. I graduated from IIT Kanpur in 2004. I have been associated with trainings related to IIT-JEE for a long time now. I c o-authored the book "Conceptual Problems in Mathematics" in 2001 and now am part of ACME Education whose subsidiaries operate as The Guidance in New Delhi and as ACME and PEETS in Bihar. Through this endeavor with Learnhub, we are trying to come up with providing students with the option to brush up their preparations in the last few months prior to taking the comeptitive engineering examinations. This portal will basically be providing you with test material and crash course options in the next few months. Based on the response that we get here, we aim to expand our base further. We believe that we have something unique to offer you as this is not a 'money making' proposition for us but an effort to give you a helping hand.

Last Updated At Dec 07, 2012
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