In the binomial expansion of {(a - b)^n},,n g | vedclass

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Question

(b) ${T_{r + 1}}\,\, = \,{}^n\,{C_r}{(a)^{n - r}}{( - \,b)^r}.$

${T_5} = {T_{4 + 1}} = {}^n{C_4}{a^{n - 4}}{( - \,b)^4}$ $ = {\,^{\,n}}{C_4}{a^{n -

Vedclass · Accepted Answer

$\frac{1}{5}(n - 4)$

Vedclass · Answer

(b) ${T_{r + 1}}\,\, = \,{}^n\,{C_r}{(a)^{n - r}}{( - \,b)^r}.$

${T_5} = {T_{4 + 1}} = {}^n{C_4}{a^{n - 4}}{( - \,b)^4}$ $ = {\,^{\,n}}{C_4}{a^{n - 4}}{b^4}$
and $6^{th}$ term

= $({T_6}) = {T_{5 + 1}} = {}^n{C_5}\,{a^{n - 5}}{( - \,b)^5}$ $ = - {\,^n}{C_5}\,{a^{n - 5}}\,{b^5}$

Since ${T_5} + {T_6} = 0$, therefore

${}^n{C_4}\,{a^{n - 4}}\,{b^4} - {}^n{C_5}\,{a^{n - 5}}\,{b^5} = 0$ ==> $\frac{{{a^{n - 4}}\,{b^4}}}{{{a^{n - 5}}\,{b^5}}} = \frac{{{}^n{C_5}}}{{{}^n{C_4}}}$

==> $\frac{a}{b} = \frac{{n!}}{{(n - 5)!\,\,5!}}\,.\,\frac{{4!\,\,(n - 4)!}}{{n!}}$ ==> $\frac{a}{b} = \frac{{n - 4}}{5}$.

In the binomial expansion of ${(a - b)^n},\,n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $\frac{a}{b}$ is equal to

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