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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.

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.

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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

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