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

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7.Binomial Theorem

normal

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

$\frac{1}{6}(n - 5)$

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

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

$\frac{6}{{(n - 5)}}$

(IIT-2001)

Solution

(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

${}^n{C_4}\,{a^{n – 4}}\,{b^4} – {}^n{C_5}\,{a^{n – 5}}\,{b^5} = 0$ ==> $\frac{{{a^{n – 4}}\,{b^4}}}{{{a^{n – 5}}\,{b^5}}} = \frac{{{}^n{C_5}}}{{{}^n{C_4}}}$

==> $\frac{a}{b} = \frac{{n!}}{{(n – 5)!\,\,5!}}\,.\,\frac{{4!\,\,(n – 4)!}}{{n!}}$ ==> $\frac{a}{b} = \frac{{n – 4}}{5}$.

Standard 11

Mathematics

If sum of the coefficient of the first, second and third terms of the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^m}$ is $46$, then the coefficient of the term that doesnot contain $x$ is :-

normal

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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

Solution

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