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Q1. Find a, b and n in the expansion of $(a + b)^n$ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Answers (1)

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As we know, the binomial expansion of $(a + b)^n$ is given by

$(a+b)^n=^nC_0a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+......^nC_nb^n$

Given in the question,

$^nC_0a^n=729.......(1)$

$^nC_1a^{n-1}b=7290.......(2)$

$^nC_2a^{n-2}b^2=30375.......(3)$

Now, dividing (1) by (2), we get:

$\Rightarrow \frac{^nC_0a^n}{^nC_1a^{n-1}b}=\frac{729}{7290}$

$\Rightarrow \frac{\frac{n!}{n!0!}}{\frac{n!}{1!(n-1)!}}\times\frac{a}{b}=\frac{729}{7290}$

$\Rightarrow\frac{(n-1)!}{n!}\times\frac{a}{b}=\frac{1}{10}$

$\Rightarrow\frac{1}{n}\times\frac{a}{b}=\frac{1}{10}$

$10a=nb......(4)$

Now, dividing (2) by (3) we get,

$\Rightarrow \frac{^nC_1a^{n-1}b}{^nC_2a^{n-2}b^2}=\frac{7290}{30375}$

$\Rightarrow \frac{\frac{n!}{1!(n-1)!}}{\frac{n!}{2!(n-2)!}}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2(n-2)!}{(n-1)!}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow \frac{2}{(n-1)}\times\frac{a}{b}=\frac{7290}{30375}$

$\Rightarrow 2\times30375\times a=7290\times b\times(n-1)$

$\Rightarrow 60750a=7290b(n-1).......(5)$

Now, from (4) and (5), we get,

$n=6,a=3\:and\:b=5$

Posted by

Pankaj Sanodiya

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