If the sum of the coefficients in the expansion of (a+b)^n is 4096, then the greatest coef…

Question

If the sum of the coefficients in the expansion of $(a+b)^n$ is 4096, then the greatest coefficient in the expansion is:

Accepted Answer

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

Solutions · 1