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If the sum of the coefficients in the expansion of $$\,{\left( {a + b} \right)^n}$$ is 4096, then the greatest coefficient in the expansion is

JEE · Math · previous-year question

  1. A.1594
  2. B.792
  3. C.924correct
  4. D.2924

Answer

C. 924

Explanation

We know, $$\,{\left( {a + b} \right)^n}$$ = $${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$$ Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable. So put $$a$$ = b = 1 $$\therefore$$ 2n = $${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$$ According to question, 2n = 4096 = 212 $$ \Rightarrow n = 12$$ So $$\,{\left( {a + b} \right)^n}$$ = $$\,{\left( {a + b} \right)^{12}}$$ Here n = 12 is even so formula for greatest term is $${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$$ For n = 12, greatest term $${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$$ $$\therefore$$ Coefficient of the greatest term = $${}^{12}{C_6}$$ = $${{12!} \over {6!6!}}$$ = 924

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