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Binomial Coefficients

If T2 / T...

Question

If $T_{2}$ / $T_{3}$ in the expansion of ${(a + b)}^{n}$ and $T_{3}$ / $T_{4}$ in the ${(a + b)}^{n + 3}$ are equal, then n=

3

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4

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5

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6

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Solution

The correct option is D $5$
$(a + b)^{n} = 1^{s t}$ Binomial Expression
$∴ T_{2}$ in $(a + b)^{n} \Rightarrow^{n} C_{1} a^{n - 1} b$
$T_{3}$ in $(a + b)^{n} \Rightarrow^{n} C_{2} a^{n - 2} b^{2}$
$T_{3}$ in $(a + b)^{n + 3} \Rightarrow^{n + 3} C_{2} a^{n + 3 - 2} b^{2}$
$T_{4}$ in $(a + b)^{n + 3} \Rightarrow^{n + 3} C_{3} a^{n + 3 - 3} b^{3}$
Acc to question
$\frac{T_{2}}{T_{3}} = \frac{T_{3}^{'}}{T_{(} 4)^{'}}$
$\frac{^{n} C_{1} a^{n - 1} b}{^{n} C_{2} a^{n - 2} b^{2}} = \frac{^{n + 3} C_{2} a^{n + 3 - 2} b^{2}}{^{n + 3} C_{3} a^{n + 3 - 3} b^{3}}$
$\frac{n \times 2}{n (n - 1)} = \frac{(n + 3)!}{(n + 1) 2!} \times \frac{n! \times 3!}{(n + 3)!}$
$\frac{2}{n - 1} = \frac{3}{n + 1}$
$2 n + 2 = 3 n - 3$
$5 = n$
$∴$ option C is correct.

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