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JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

  • question_answer
    If the coefficients of second, third and fourth term in the expansion of \[{{(1+x)}^{2n}}\] are in A.P., then \[2{{n}^{2}}-9n+7\] is equal to    [AMU 2001; MP PET 2004]

    A) - 1

    B) 0

    C) 1

    D) 3/2

    Correct Answer: B

    Solution :

    \[{{T}_{2}}={}^{2n}{{C}_{1}}\,\,x\], \[{{T}_{3}}={}^{2n}{{C}_{2}}\,\,{{x}^{2}}\],  \[{{T}_{4}}={}^{2n}{{C}_{3}}\,\,{{x}^{3}}\] Coefficient of T2, T3, T4 are in A.P. Þ \[2.{}^{2n}{{C}_{2}}={}^{2n}{{C}_{1}}+{}^{2n}{{C}_{3}}\] Þ \[2\frac{2n!}{2\,!\,(2n-2)\,!}=\frac{2n!}{(2n-1)\,!}+\frac{2n!}{3\,!\,(2n-3)\,!}\] Þ \[\frac{2\,.\,2n(2n-1)}{2}=2n+\frac{\,2n(2n-1)(2n-2)}{6}\] Þ \[n(2n-1)=n+\frac{(n)(2n-1)(2n-2)}{6}\] Þ \[6(2{{n}^{2}}-n)=6n+4{{n}^{3}}-6{{n}^{2}}+2n\] Þ \[6n(2n-1)=2n(2{{n}^{2}}-3n+4)\] Þ \[6n-3=2{{n}^{2}}-3n+4\] Þ \[0=2{{n}^{2}}-9n+7\]Þ \[2{{n}^{2}}-9n+7=0\].


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