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Binomial Coefficients

If t2/t3 in...

Question

If $\frac{t_{2}}{t_{3}}$ in the expression of $(a + b)$ and $\frac{t_{3}}{t_{4}}$ in the expression of $(a + b)^{n + 3}$ are equal, then find $n$

A

3

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B

4

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C

5

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D

6

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Solution

The correct option is C $5$
$(a + b)^{n}$
$t_{2} =^{n} c_{1} a^{n - 1} b^{1}$
$t_{3} =^{n} c_{2} a^{n - 2} b^{2}$
$\frac{t_{2}}{t_{3}} = \frac{^{n} c_{1}}{^{n} c_{2}} . \frac{a^{n - 1} . b}{a^{n - 2} . b} = \frac{^{n} c_{1}}{^{n} c_{2}} . \frac{a}{b}$
$(a + b)^{n + 3}$
$t_{3}^{‘} =^{n + 3} c_{2} (a)^{n + 1} (b)^{2}$
$t_{4}^{‘} =^{n + 3} c_{3} (a)^{n} (b)^{3}$
$\frac{t_{3}^{‘}}{t_{4}^{‘}} = \frac{^{n + 3} c_{2}}{^{n + 3} c_{3}} . \frac{a^{n + 1}}{a^{n}} . \frac{b^{2}}{b^{3}}$
$= \frac{^{n + 3} c_{2}}{^{n + 3} c_{3}} \frac{b^{2}}{b^{3}}$
$\frac{t_{2}}{t_{3}} = \frac{t_{3}^{‘}}{t_{4}^{‘}}$
$\frac{^{n} c_{1}}{^{n} c_{2}} . \frac{a}{b} = \frac{^{n + 3} c_{2}}{^{n + 3} c_{3}} . \frac{a}{b}$
$\frac{^{n} c_{1}}{^{n} c_{2}} = \frac{^{n + 3} c_{2}}{^{n + 3} c_{3}}$
$\frac{\frac{n!}{1! (n - 1)!}}{\frac{n!}{2! (n - 2)!}} = \frac{\frac{(n + 3)!}{2! (n + 1)!}}{\frac{(n + 3)!}{3! (n)!}}$
$\frac{2}{n - 1} = \frac{3 n!}{(n + 1) (n)!}$
$2 n + 2 = 3 n - 3$
$- n = - 5$
$\Rightarrow n = 5$

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